Solutions to Supplementary

Chapter 5 Solutions to Supplementary Check for Understanding Problems Moles and Molar Mass 1. Indicate the appropriate quantity for each of the following. a) b) c) d) e) Answers: A mole of N atoms contains _________________ atoms. A mole of N2 molecules contains _________________ molecules. A mole of N2 molecules contains _________________ atoms. A mole of N atoms has a mass of _________________ grams. A mole of N2 molecules has a mass of _________________ grams. a) 6.022 x 1023 atoms b) 6.022 x 1023 molecules c) 1.2044 x 1024 atoms d) 14.01 g e) 28.02 g Solutions a) A mole of anything always contains 6.022 x 1023 items of that material. b) A mole of anything always contains 6.022 x 1023 items of that material. c) Since a mole of N2 molecules contains 6.022 x 1023 molecules of N2 and there are 2 atoms of N per molecule of N2, the total number of N atoms is given by: 6.022 x10 23 N 2 molecules 2 atoms N 1.2044 10 24 atoms N x = mol N 2 1 molecule N 2 mol N 2 d) The mass in grams of a mole of atoms of any element (its molar mass) is numerically equally to the weighted average atomic mass in atomic mass units. Since the atomic weight of nitrogen is 14.01, its weighted atomic mass is 14.01 u and a mole of nitrogen atoms weighs 14.01 g. S.5.5 S.5.6 e) CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The mass in grams of a mole of N2 molecules (its molar mass) is obtained by summing the molar masses of the atoms in the chemical formula. Since the molar mass of N is 14.01 g/mol (see part d), the molar mass of N2 is: 14.01g 2 mol N 28.01g x = mol N 2 mol N 1mol N 2 2. a) What is the mass of 725 sodium atoms in atomic mass units? Answer: 1.67 x 103 u Solution The numerical value of the atomic weight of sodium (22.99) refers to the mass of a single sodium atom in atomic mass units. Applying this yields: 725 Na atoms x 22.99 u = 1.67 x103 u Na atom b) What is the mass of 725 sodium atoms in grams? Answer: 2.77 x 10-20 g Solution What we know: number of Na atoms Desired answer: g Na The solution map for this calculation is: atom Na ö mol Na ö g Na The conversion factor needed in the first step is the Avogadro constant expressed in the form 1mol Na . 6.022 x1023 Na atoms The conversion factor needed in the second step is the molar mass of sodium. The numerical value for the molar mass is obtained from the atomic weight of sodium (22.99) and is expressed 22.99 g Na in the form . 1mol Na S.5.7 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Putting these together yields: 725 atoms Na x 3. 1 mol Na 22.99 g Na x = 2.77 x10-20 g Na 23 1 mol Na 6.022 x10 atoms Na How many atoms of an element are present in a sample of that element if the sample has a mass in grams numerically equal to the atomic weight of the element? Answer: 6.022 x 1023 atoms Solution A mass in grams numerically equal to the atomic weight of an element is its molar mass and thus contains one mole of atoms of that element. Therefore, this sample will contain 6.022 x 1023 atoms. 4. Blackboard chalk is mostly calcium sulfate. How would you determine how many moles of calcium sulfate it takes to write your name in chalk on a blackboard? Solution The moles of a pure substance can be determined from the mass of the substance and its molar mass. The mass of the chalk used can be obtained by weighing the chalk before and after writing your name and calculating the difference between the masses. The molar mass of calcium sulfate can be obtained from its chemical formula (CaSO4). The calculations will be: g CaSO4 used = initial mass of chalk - final mass of chalk g CaSO 4 used x 5. 1mol CaSO 4 = mol CaSO 4 used 136.15 g CaSO 4 What mass of zinc metal contains the same number of atoms as 16.1 grams of silver? Answer: 9.76 g Solution What we know: g Ag; number of Ag atoms equals number of Zn atoms Desired answer: g Zn S.5.8 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The solution map for this calculation is: g Ag ö mol Ag ö mol Zn ö g Zn The conversion factor needed in the first step is the molar mass of Ag in the form 1mol Ag . 107.9 g Ag The problem indicates that the number of atoms of Zn is the same as the number of atoms of Ag. Therefore, mol Zn = mol Ag. This relationship can be applied as the conversion factor in the 1mol Zn second step in the form . 1mol Ag The conversion factor needed in the last step is the molar mass of Zn expressed in the form 65.39 g Zn . mol Zn Putting these together yields: 16.1 g Ag x 6. 1 mol Ag 1 mol Zn 65.39 g Zn x x = 9.76 g Zn 1 mol Zn 107.9 g Ag 1 mol Ag One atom of an element is found to weigh 2.107 x 10-22 g. What is the atomic weight of this element? Answer: 126.9 Solution What we know: g/atom Desired answer: atomic weight of element We know that the atomic weight of this element is numerically equal to the molar mass of this element. The solution map for calculating the molar mass is: g atom ö g mol The conversion factor needed is the Avogadro constant expressed in the form 6.022 x1023 atoms . mol S.5.9 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Applying this yields: 2.107 x10-22 g 6.022 x10 23 atoms 126.9 g x = atom mol mol Since the molar mass of this element is 126.9 g/mol, then the atomic weight of the element is 126.9. 7. Which has the larger mass, 1.0 mmol of calcium or 1.5 mmol of sulfur? Justify your choice. Answer: 1.5 mmol sulfur Solution The most direct way to make this determination is to calculate the mass of each and compare. The general solution map for this calculation is: mmol element ö mol element ö g element The conversion factor needed in the first step is that between mmol and mol in the form 10-3 mol . 1mmol The conversion factor needed in the second step is the molar mass of the element expressed in g the form . mol Applying these yields: 1.0 mmol Ca x 10-3 mol Ca 40.08g Ca = 0.040 g Ca x mol Ca 1 mmol Ca 10-3 molS 32.07 g S 1.5 mmolS x = 0.048g S x molS 1 mmolS 8. ³ larger mass Which has the larger number of atoms, 0.045 μg of nickel or 0.032 μg of potassium? Justify your choice. Answer: 0.032 μg potassium S.5.10 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Solution The most direct way to make this determination is to calculate the moles of each and compare. The general solution map for this calculation is: μg element ö g element ö mol element The conversion factor needed in the first step is that between μg and g in the form 10-6 g . 1μg The conversion factor needed in the second step is the molar mass of the element expressed in mol the form . g Applying these yields: 0.045 μg Ni x 0.032 μg K x 9. 10-6 g Ni 1mol Ni x = 7.7 x10-10 mol Ni 58.69 g Ni 1 μg Ni 10-6 g K 1mol K x = 8.2 x10-10 mol K 39.10 g K 1 μg K ³ larger mol Calculate the molar mass for each of the following compounds. a) potassium hydrogen phosphate b) Pb(C2H3O2)2 Answers: a) 174.18 g/mol b) 325.29 g/mol Solutions a) What we know: potassium hydrogen phosphate; atomic weights of elements Desired answer: g potassium hydrogen phosphate/mol S.5.11 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The chemical formula for potassium hydrogen phosphate is K2HPO4. The formula indicates that in 1 mole of potassium hydrogen phosphate there are 2 moles of potassium, 1 mole of hydrogen, 1 mole of phosphorus and 4 moles of oxygen. From the periodic table we see that 1 mole of potassium atoms weighs 39.10 g, 1 mole of hydrogen atoms weighs 1.008 g, 1 mole of phosphorus atoms weighs 30.97 g and 1 mole of oxygen atoms weighs 16.00 g. Thus, one mole of K2HPO4 will weigh: 2K 2 mol K x 39.10 g = mol K 1H 1 mol H x 1.008g = mol H 1P 1 mol P x 30.97 g = mol P 30.97 g 4O 4 mol O x 16.00 g = 1 mol O 64.00 g 78.20 g 1.008g 174.18 g The molar mass of potassium hydrogen phosphate is 174.18 g/mol. b) What we know: Pb(C2H3O2)2; atomic weights of elements Desired answer: g Pb(C2H3O2)2/mol The chemical formula indicates that in 1 mole of lead(II) acetate there are 1 mole of lead, 4 moles of carbon, 6 moles of hydrogen and 4 moles of oxygen. From the periodic table we see that 1 mole of lead atoms weighs 207.2 g, 1 mole of carbon atoms weighs 12.01 g, 1 mole of hydrogen atoms weighs 1.008 g and 1 mole of oxygen atoms weighs 16.00 g. Thus, one mole of K2HPO4 will weigh: S.5.12 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS 1 Pb 1 mol Pb x 207.2 g = 207.2 g 1 mol Pb 4C 4 mol C x 12.01 g = 1 mol C 6H 6 mol H x 1.008g = 1 mol H 4O 4 mol O x 16.00 g = 1 mol O 48.04 g 6.048g 64.00 g 325.29 g The molar mass of lead(II) acetate is 325.29 g/mol. 10. Calculate the number of moles of compound in each of the following samples. a) 2.239 g C2H5OH b) 63.1 ng sulfur trioxide c) 1.48 x 102 kg potassium permanganate Answers: a) 0.04860 mol b) 7.88 x 10-10 mol c) 9.36 x 102 mol Solutions a) What we know: g C2H5OH Desired answer: mol C2H5OH The solution map for this calculation is g C2H5OH ö mol C2H5OH S.5.13 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The conversion factor needed is the molar mass of C2H5OH. From the periodic table we can get the molar masses of carbon, hydrogen and oxygen and add them as follows. 2(12.01 g) C + 6(1.008 g) H + 16.00 g O = 46.07 g. Thus the molar mass of C2H5OH is 46.07 g/mol. This is 1mol C 2 H 5 OH used in the form to convert units properly. Applying this yields: 46.07 g C 2 H 5 OH 2.239 g C2 H 5 OH x b) 1mol C 2 H 5 OH = 0.04860 mol C 2 H 5 OH 46.07 g C 2 H 5 OH What we know: ng sulfur trioxide Desired answer: mol sulfur trioxide The solution map for this calculation is ng sulfur trioxide ö g sulfur trioxide ö mol sulfur trioxide The formula for sulfur trioxide is SO3. The conversion factor needed in the first step is that 10-9 g between ng and g in the form . The conversion factor needed in the second step is the 1ng molar mass of SO3. From the periodic table we can get the molar masses of sulfur and oxygen and add them as follows. 32.07 g S + 3(16.00 g) O = 80.07 g. Thus the molar mass of SO3 is 1molSO3 80.07 g/mol. This is used in the form to convert units properly. 80.07 g SO3 Putting these together yields: 10-9 g SO3 1molSO3 x 63.1 ng SO3 x = 7.88 x10-10 molSO3 80.07 g SO3 1 ng SO3 c) What we know: kg potassium permanganate Desired answer: mol potassium permanganate The solution map for this calculation is kg potassium permanganate ö g potassium permanganate ö mol potassium permanganate S.5.14 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The formula for potassium permanganate is KMnO4. The conversion factor needed in the first 103 g step is that between kg and g in the form . The conversion factor needed in the second step 1kg is the molar mass of KMnO4. From the periodic table we can get the molar masses of potassium, manganese and oxygen and add them as follows. 39.10 g K + 54.94 g Mn + 4(16.00 g) O = 158.04 g. Thus the molar mass of KMnO4 is 158.04 g/mol. This is used in the 1mol KMnO 4 form to convert units properly. 158.04 g KMnO 4 Putting these together yields: 103 g KMnO 4 1mol KMnO 4 1.48 x10 kg KMnO 4 x = 9.36 x102 mol KMnO 4 x 158.04 g KMnO 4 1 kg KMnO 4 2 11. How many CO molecules are present in 18.4 metric tons of carbon monoxide? One metric tons equals 1000 kg. Answer: 3.96 x 1029 molecules Solution What we know: metric tons CO Desired answer: number of molecules CO The solution map for this calculation is metric tons CO ö kg CO ö g CO ö mol CO ö molecules CO The conversion factor needed in the first step is that between metric tons and kg in the form 1000 kg . The conversion factor needed in the second step is that between kg and g in the 1metric ton 103 g . The conversion factor needed in the third step is the molar mass of CO. From the 1kg periodic table we can get the molar masses of carbon and oxygen and add them as follows. 12.01 g C + 16.00 g O = 28.01 g. Thus the molar mass of CO is 28.01 g/mol. This is used in the 1mol CO form to convert units properly. 28.01g CO form S.5.15 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The conversion factor needed in the last step is the Avogadro constant in the form 6.022 x10 23 molecules CO . mol CO Putting these together yields: 18.4 metric tons CO x 12. 1000 kg CO 103 g CO 1 mol CO 6.022 x1023 molecules CO x x x = 3.96 1029 molecules CO 1 mol CO 1 metric ton CO 1 kg CO 28.01 g CO Calculate the mass in grams of each of the following samples. a) 9.44 mol copper(II) sulfate b) 7.11 mmol Li2CO3 Answers: a) 1.51 x 103 g b) 0.427 g Solutions a) What we know: mol copper(II) sulfate Desired answer: g copper(II) sulfate The solution map for this calculation is mol copper(II) sulfate ö g copper(II) sulfate The formula for copper(II) sulfate is CuSO4. The conversion factor needed is the molar mass of CuSO4. From the periodic table we can get the molar masses of copper, sulfur and oxygen and add them as follows. 63.55 g Cu + 32.07 g S + 4(16.00 g) O = 159.62 g. Thus the molar mass 159.62 g CuSO 4 of CuSO4 is 159.62 g/mol. This is used in the form to convert units properly. 1mol CuSO 4 Applying this yields: 9.44 mol CuSO 4 x 159.62 g CuSO 4 = 1.51x103 g CuSO 4 1 mol CuSO 4 S.5.16 b) CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS What we know: mmol Li2CO3 Desired answer: g Li2CO3 The solution map for this calculation is mmol Li2CO3 ö mol Li2CO3 ö g Li2CO3 The conversion factor needed in the first step is that between mmol and mol in the form 10-3 mol . The conversion factor needed in the second step is the molar mass of Li2CO3. From 1mmol the periodic table we can get the molar masses of lithium, carbon and oxygen and add them as follows. 2(6.941) g Li + 12.01 g C + 3(16.00 g) O = 60.01 g. Thus the molar mass of Li2CO3 is 60.01g Li 2 CO3 60.01 g/mol. This is used in the form to convert units properly. 1mol Li 2 CO3 Putting these together yields: 7.11 mmol Li 2 CO3 x 13. 10-3 mol Li 2 CO3 60.01g Li 2 CO3 x = 0.427 g Li 2 CO3 1 mol Li 2 CO3 1 mol Li 2 CO3 Calculate the moles of sulfur atoms in each of the following samples. a) 4.63 g sodium thiosulfate b) 5.81 μg Na2S Answers: a) 0.0586 mol b) 7.44 x 10-8 mol Solutions a) What we know: g sodium thiosulfate Desired answer: mol sulfur atoms The solution map for this calculation is: g sodium thiosulfate ö mol sodium thiosulfate ö mol S S.5.17 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The formula for sodium thiosulfate is Na2S2O3. The conversion factor needed in the first step is the molar mass of Na2S2O3. From the periodic table we can get the molar masses of sodium, sulfur and oxygen and add them as follows. 2(22.99 g) Na + 2(32.07 g) S + 3(16.00 g) O = 158.12 g. Thus the molar mass of Na2S2O3 is 158.12 g/mol. This is needed in the form 1mol Na 2S2 O3 . 158.12 g Na 2S2 O3 The formula of Na2S2O3 indicates that in 1 mole of Na2S2O3 there are 2 moles of sulfur atoms. This relationship can be applied as the conversion factor in the second step in the form 2 molS . 1mol Na 2S2 O3 Putting these together yields: 4.63 g Na 2S2 O3 x b) 1 mol Na 2S2 O3 2 molS x = 0.0586 molS 158.12 g Na 2S2 O3 1 mol Na 2S2 O3 What we know: μg Na2S Desired answer: mol S atoms The solution map for this calculation is: μg Na2S ö g Na2S ö mol Na2S ö mol S 10-6 g The conversion factor needed in the first step is that between μg and g in the form . The 1μg conversion factor needed in the second step is the molar mass of Na2S. From the periodic table we can get the molar masses of sodium and sulfur and add them as follows. 2(22.99 g) Na + 32.07 g S = 78.05 g. Thus the molar mass of Na2S is 78.05 g/mol. This is needed in the form 1mol Na 2S . 78.05g Na 2S The formula of Na2S indicates that in 1 mole of Na2S there is 1 mole of sulfur atoms. This relationship can be applied as the conversion factor in the last step in the form 1molS . 1mol Na 2S Putting these together yields: 10-6 g Na 2S 1 mol Na 2S 1molS x 5.81 μg Na 2S x = 7.44 x10-8 molS x 1 μg Na 2S 78.05 g Na 2S 1 mol Na 2S S.5.18 14. CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Calculate the number of carbon atoms in a 3.92-g sample of C6H4Cl2. Answer: 9.64 x 1022 atoms Solution What we know: g C6H4Cl2 Desired answer: number of C atoms The solution map for this calculation is: g C6H4Cl2 ö mol C6H4Cl2 ö mol C ö atoms C The conversion factor needed in the first step is the molar mass of C6H4Cl2. From the periodic table we can get the molar masses of carbon, hydrogen and chlorine and add them as follows. 6(12.01 g) C + 4(1.008 g) H + 2(35.45 g) Cl = 146.99 g. Thus the molar mass of C6H4Cl2 is 1mol C6 H 4 Cl2 146.99 g/mol. This is needed in the form . 146.99 g C6 H 4 Cl 2 The formula of C6H4Cl2 indicates that in 1 mole of C6H4Cl2 there are 6 moles of carbon atoms. This relationship can be applied as the conversion factor in the second step in the form 6 mol C . 1mol C6 H 4 Cl2 The conversion factor needed in the last step is the Avogadro constant in the form 6.022 x10 23 atoms C . 1mol C Putting these together yields: 1mol C6 H 4 Cl2 6 mol C 6.022 x1023 atoms C 3.92 g C6 H 4 Cl2 x = 9.64 x1022 atoms C x x 1 mol C 146.99 g C6 H 4 Cl2 1 mol C6 H 4 Cl 2 15. How many moles of oxygen atoms are present in 4.40 mmol calcium phosphate? Answer: 0.0352 mol S.5.19 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Solution What we know: mmol calcium phosphate Desired answer: mol O The solution map for this calculation is: mmol calcium phosphate ö mol calcium phosphate ö mol O The formula for calcium phosphate is Ca3(PO4)2. The conversion factor needed in the first step 10-3 mol is that between mmol and mol in the form . 1mmol The formula of Ca3(PO4)2 indicates that in 1 mole of Ca3(PO4)2 there are 8 moles of oxygen atoms. This relationship can be applied as the conversion factor in the second step in the form 8 mol O . 1mol Ca 3 (PO 4 ) 2 Putting these together yields: 10-3 mol Ca 3 (PO 4 ) 2 8 mol O x 4.40 mmol Ca 3 (PO 4 ) 2 x = 0.0352 mol O 1 mol Ca 3 (PO 4 ) 2 1 mmol Ca 3 (PO 4 ) 2 Mass Percent 1. Calculate the mass percent of each element in the following compounds. a) barium chloride b) sodium sulfate Answers: a) 65.95 % Ba and 34.05 % Cl b) 32.37 % Na, 22.58 % S and 45.05 % O Solutions a) What we know: barium chloride; atomic weights of elements Desired answer: mass % of each element present S.5.20 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The formula for barium chloride is BaCl2. Since the mass percent values apply to any sample of BaCl2, it is convenient to consider one mole of this compound. From the periodic table we can get the molar masses of barium and chlorine and add them as follows. 137.3 g Ba + 2(35.45 g) Cl = 208.2 g. Thus the molar mass of BaCl2 is 208.2 g/mol, so 208.2 g represents the total mass of material. The formula indicates that 1 mole of BaCl2 contains 1 mole of barium. Therefore the mass of barium (=component mass) present is 137.3 g. These values are used to obtain the mass percent for barium. mass % Ba = 137.3g Ba x 100 = 65.95% 208.2 g BaCl2 Since chlorine is the only other element present in the compound, the mass % Cl = 100.00 % 65.95 % = 34.05 %. b) What we know: sodium sulfate; atomic weights of elements Desired answer: mass % of each element present The formula for sodium sulfate is Na2SO4. Since the mass percent values apply to any sample of Na2SO4, it is convenient to consider one mole of this compound. From the periodic table we can get the molar masses of sodium, sulfur and oxygen and add them as follows. 2(22.99 g) Na + 32.07 g S + 4(16.00 g) O = 142.05 g. Thus the molar mass of Na2SO4 is 142.05 g/mol, so 142.05 g represents the total mass of material. The formula indicates that 1 mole of Na2SO4 contains 2 moles of sodium. Therefore the mass of sodium (=component mass) present is 2(22.99 g) = 45.98 g. These values are used to obtain the mass percent for sodium. mass % Na = 45.98g Na x 100 = 32.37 % 142.05g Na 2SO 4 The formula indicates that 1 mole of Na2SO4 contains 1 mole of sulfur. Therefore the mass of sulfur (=component mass) present is 32.07 g. These values are used to obtain the mass percent for sulfur. mass %S = 32.07 g S x 100 = 22.58% 142.05g Na 2SO 4 Since oxygen is the only other element present in the compound, the mass % O = 100.00 % 32.37 % - 22.58% = 45.05 %. S.5.21 2. CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Which of the following compounds contains the largest mass percent of nitrogen? Justify your choice. a) NH4NO3 Answer: b) HNO3 c) N2O4 d) Al(NO3)3 NH4NO3 Solution The most direct way to answer this is to determine the mass percent N in each compound. This is calculated from the mass of one mole of a compound and the mass of nitrogen present in one mole of that compound. The required data are shown below. Compound Mass of one mole of compound Mass of N in one mole of compound Mass % N NH4NO3 80.05 g 28.02 g 35.00 % HNO3 63.02 g 14.01 g 22.23 % N2O4 92.02 28.02 g 30.45 % Al(NO3)3 213.01 g 42.03 g 19.72 % Thus, NH4NO3 has the largest mass % N. 3. In a particular molecular compound the mass percent sulfur is 50.0% and the mass percent oxygen is 50.0%. What is the ratio of oxygen atoms to sulfur atoms in a molecule of this compound? Answer: 2 O atoms/S atom Solution What we know: mass % of each element present; atomic weights of elements Desired answer: number of O atoms/S atom in compound The ratio of O atoms to S atoms is the same as the ratio of mol O to mol S. In order to obtain the moles of each element, choose an arbitrary mass of the compound (100 g simplifies the math) and use the following solution map: g compound ö g element ö mol element S.5.22 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The conversion factor for the first step is the mass percent of the element used in the form g element . The conversion factor for the second step is the molar mass of the element in 100 g compound the form mol element . g element Applying these to each element yields: 100 g compound x 50.0 g O 1mol O x = 3.12 mol O 100 g compound 16.00 g O 100 g compound x 50.0 g S 1molS x = 1.56 molS 100 g compound 32.07 g S The mole ratio of O to S is 3.12 mol O/1.56 mol S = 2 which is the same as the atom ratio. 4. If a type of stainless steel contains 18% chromium by mass, how many moles of chromium are present in a bar of this material weighing 1.5 kg? Answer: 5.2 mol Solution What we know: mass of sample; mass % Cr Desired answer: mol Cr The solution map for this calculation is: kg sample ö g sample ö g Cr ö mol Cr 103 g The conversion factor needed in the first step is that between kg and g in the form . 1kg The conversion factor for the second step is the mass percent of Cr used in the form 18g Cr . 100 g sample The conversion factor for the last step is the molar mass of Cr in the form 1mol Cr . 52.00 g Cr S.5.23 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Putting these together yields: 1.5 kg sample x 103 g sample 18 g Cr 1mol Cr x x = 5.2 mol Cr 1 kg sample 100 g sample 52.00 g Cr Stoichiometric Calculations (mole-to-mole) 1. Balance the following equation and state the meaning of the equation in terms of individual units of reactants and products and in terms of moles of reactants and products. Al(s) + MnO2(s) ÷ Mn(s) + Al2O3(s) Answers: 4Al(s) + 3MnO2(s) ÷ 3Mn(s) + 2Al2O3(s) 4 atoms 3 formula units 4 mol 3 mol 3 atoms 3 mol 2 formula units 2 mol Solution To balance this equation, place a coefficient of 3 in front of MnO2 and a coefficient of 2 in front of Al2O3 to balance oxygen atoms. This requires a coefficient of 3 in front of Mn to balance manganese atoms. Finally, a coefficient of 4 in front of Al will balance aluminum atoms. The balanced equation is: 4Al(s) + 3MnO2(s) ÷ 3Mn(s) + 2Al2O3(s) At the most fundamental level this equation suggests that 4 atoms of aluminum react with 3 formula units of MnO2 to produce 3 atoms of Mn and 2 formula units of Al2O3. Multiplying each reactant amount and each product amount by the Avogadro constant yields 4 mol Al reacting with 3 mol MnO2 to form 3 mol Mn and 2 mol Al2O3. 2. How many moles of CO2 are needed to react completely with 0.675 mol LiOH? LiOH(aq) + CO2(g) ÷ Li2CO3(aq) + H2O(l) Answer: 0.338 mol (unbalanced) S.5.24 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Solution What we know: mol LiOH; equation relating CO2 and LiOH Desired answer: mol CO2 The solution map for this calculation is: mol LiOH ö mol CO2 The conversion factor needed is the mole ratio for these two substances from the balanced equation. To balance this equation place a coefficient of 2 in front of LiOH. The balanced equation is: 2LiOH(aq) + CO2(g) ÷ Li2CO3(aq) + H2O(l) Applying the mole ratio yields: 0.675 mol LiOH 3. 1mol CO 2 = 0.338 mol CO 2 2 mol LiOH Given the reaction 4FeS(s) + 7O2(g) ÷ 2Fe2O3(s) + 4SO2(g) how many moles of O2 are needed to: a) produce 0.693 mol Fe2O3? b) react completely with 9.14 mol FeS? c) form 1.51 mol SO2? Answers: a) 2.43 mol b) 16.0 mol c) 2.64 mol Solutions a) What we know: mol Fe2O3; balanced equation relating O2 and Fe2O3 Desired answer: mol O2 S.5.25 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The solution map for this problem is: mol Fe2O3 ö mol O2 The conversion factor needed is the mole ratio for these two substances from the balanced 7 mol O 2 equation in the form . 2 mol Fe 2 O3 Applying this yields: 0.693 mol Fe 2 O3 x b) 7 mol O 2 = 2.43mol O 2 2 mol Fe 2 O3 What we know: mol FeS; balanced equation relating O2 and FeS Desired answer: mol O2 The solution map for this problem is: mol FeS ö mol O2 The conversion factor needed is the mole ratio for these two substances from the balanced 7 mol O 2 equation in the form . 4 mol FeS Applying this yields: 9.14 mol FeS x c) 7 mol O 2 = 16.0 mol O 2 4 mol FeS What we know: mol SO2; balanced equation relating O2 and SO2 Desired answer: mol O2 The solution map for this problem is: mol SO2 ö mol O2 The conversion factor needed is the mole ratio for these two substances from the balanced 7 mol O 2 equation in the form . 4 molSO 2 S.5.26 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Applying this yields: 1.51 molSO 2 x 7 mol O 2 = 2.64 mol O 2 4 molSO 2 Stoichiometric Calculations (mole-to-mass & mass-to-mole) 1. How many moles of each product can be formed from the decomposition of 1.00 g of the rocket fuel hydrazine (N2H4)? 3N2H4(l) ÷ 4NH3(g) + N2(g) Answers: 0.0416 mol NH3 and 0.0104 mol N2 Solution What we know: g N2H4; balanced equation relating NH3, N2 and N2H4 Desired answer: mol NH3 and mol N2 The solution maps for these calculations are: g N2H4 ö mol N2H4 ö mol NH3 g N2H4 ö mol N2H4 ö mol N2 In each case, the conversion factor needed in the first step is the molar mass of N2H4. From the periodic table we can get the molar masses of nitrogen and hydrogen and add them as follows. 2(14.01 g) N + 4(1.008 g) H = 32.05 g. Thus, the molar mass of N2H4 is 32.05 g/mol. It is used 1mol N 2 H 4 in the form in order to cancel units properly. 32.05g N 2 H 4 The conversion factor needed in the second step is the mole ratio for the particular product and 4 mol NH 3 1mol N 2 and N2H4: 3mol N 2 H 4 3mol N 2 H 4 S.5.27 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Putting these together yields: 1.00 g N 2 H 4 x 4 mol NH 3 1 mol N 2 H 4 = 0.0416 mol NH 3 x 32.05 g N 2 H 4 3 mol N 2 H 4 1.00 g N 2 H 4 x 1 mol N 2 H 4 1mol N 2 = 0.0104 mol N 2 x 32.05 g N 2 H 4 3 mol N 2 H 4 Note that the number of moles of NH3 is four(4) times that of N2 as suggested by the balanced equation. 2. How many moles of oxygen gas are needed for the complete combustion of 19.6 g acetylene, C2H2(g)? Answer: 1.88 mol Solution What we know: g C2H2 Desired answer: mol O2 First, a balanced equation is needed for the reaction between C2H2 and O2. Since this is a combustion reaction, the products are CO2(g) and H2O(l). C2H2(g) + O2(g) ÷ CO2(g) + H2O(l) Notice that the hydrogen atoms are balanced so first, balance the carbon atoms. Since there are 2 C atoms on the reactant side and only 1 C atom on the product side, a coefficient of 2 in front of CO2 will balance carbon atoms. C2H2(g) + O2(g) ÷ 2CO2(g) + H2O(l) (H and C atoms balanced) Finally, balance the oxygen atoms. Since there are 2 O atoms on the reactant side and 5 O atoms on the product side, multiplying O2 by 2½, or 5/2, will balance oxygen atoms. C2H2(g) + 5/2O2(g) ÷ 2CO2(g) + H2O(l) (all atoms balanced) S.5.28 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The fractional coefficient in front of O2 can be converted to the smallest whole number by multiplying by 2. This requires that all other coefficients be multiplied by 2 in order to retain the atom balance. The resulting balanced equation is: 2C2H2(g) + 5O2(g) ÷ 4CO2(g) + 2H2O(l) (all atoms balanced using whole-number coefficients) The solution map for this problem is: g C2H2 ö mol C2H2 ö mol O2 The conversion factor needed in the first step is the molar mass of C2H2. From the periodic table we can get the molar masses of carbon and hydrogen and add them as follows. 2(12.01 g) C + 2(1.008 g) H = 26.04 g. Thus, the molar mass of C2H2 is 26.04 g/mol. It is used in the form 1mol C 2 H 2 in order to cancel units properly. 26.04 g C 2 H 2 The conversion factor needed in the second step is the mole ratio for these two substances from 5 mol O 2 the balanced equation in the form . 2 mol C 2 H 2 Putting these together yields: 19.6 g C2 H 2 x 3. 1 mol C 2 H 2 5 mol O 2 = 1.88 mol O 2 x 26.04 g C 2 H 2 2 mol C 2 H 2 How many kilograms of Li2O are needed to react completely 4.17 x 103 mol H2O? Li2O(s) + H2O(g) ÷ 2LiOH(s) Answer: 125 kg Solution What we know: mol H2O; balanced equation relating Li2O and H2O Desired answer: kg Li2O The solution map for this problem is: mol H2O ö mol Li2O ö g Li2O ö kg Li2O S.5.29 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The conversion factor needed in the first step is the mole ratio for these two substances from the 1mol Li 2 O balanced equation in the form . 1mol H 2 O The conversion factor needed in the second step is the molar mass of Li2O. From the periodic table we can get the molar masses of lithium and oxygen and add them as follows. 2(6.941 g) Li + 16.00 g O = 29.88 g. Thus, the molar mass of Li2O is 29.88 g/mol. It is used in the form 29.88g Li 2 O in order to cancel units properly. 1mol Li 2 O The conversion factor needed in the last step is that between g and kg in the form 1kg . 103 g Putting these together yields: 4.17 x103 mol H 2 O x 4. 1 mol Li 2 O 29.88 g Li 2 O 1kg Li 2 O = 125 kg Li 2 O x x 3 10 g Li 2 O 1 mol H 2 O 1 mol Li 2 O Carbon dioxide is produced in the reaction H3PO4(aq) + MgCO3(s) ÷ Mg3(PO4)2(s) + CO2(g) + H2O(l) (unbalanced) How many grams of MgCO3 are needed to produce 14.8 moles of CO2? Answer: 1.25 x 103 g Solution What we know: mol CO2; equation relating MgCO3 and CO2 Desired answer: g MgCO3 First, a balanced equation is needed. To balance this equation, place a coefficient of 2 in front of MgCO3 to balance magnesium atoms. Then place a coefficient of 2 in front of H3PO4 to balance PO4 groups. Next, a coefficient of 3 in front of CO2 will balance carbon atoms. Finally, a coefficient of 3 in front of H2O will balance hydrogen and oxygen atoms. The balanced equation is: 2H3PO4(aq) + 3MgCO3(s) ÷ Mg3(PO4)2(s) + 3CO2(g) + 3H2O(l) The solution map for this problem is: mol CO2 ö mol MgCO3 ö g MgCO3 S.5.30 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The conversion factor needed in the first step is the mole ratio for these two substances from the 3mol MgCO3 balanced equation in the form . 3mol CO 2 The conversion factor needed in the second step is the molar mass of MgCO3. From the periodic table we can get the molar masses of magnesium, carbon and oxygen and add them as follows. 24.31 g Mg + 12.01 g C + 3(16.00 g) O = 84.32 g. Thus, the molar mass of MgCO3 is 84.32 84.32 g MgCO3 g/mol. It is used in the form in order to cancel units properly. 1mol MgCO3 Putting these together yields: 14.8 mol CO 2 x 3 mol MgCO3 84.32 g MgCO3 x = 1.25 x103 g MgCO3 1 mol MgCO3 3 mol CO 2 Stoichiometric Calculations (mass-to-mass) 1. How many grams of sulfur can react with 1.79 g of copper according to the following equation? Cu(s) + S(s) ÷ CuS(s) Answer: 0.903 g Solution What we know: g Cu; balanced equation relating S and Cu Desired answer: gS The solution map for this problem is: g Cu ö mol Cu ö mol S ö g S The conversion factor needed in the first step is the molar mass of Cu. From the periodic table 1mol Cu we see that the molar mass of copper is 63.55 g/mol. It is used in the form in order 63.55g Cu to cancel units properly. The conversion factor needed in the second step is the mole ratio for these two substances from 1molS the balanced equation in the form . 1mol Cu S.5.31 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The conversion factor needed in the last step is the molar mass of S. From the periodic table we 32.07 g S see that the molar mass of sulfur is 32.07 g/mol. It is used in the form in order to 1molS cancel units properly. Putting these together yields: 1.79 g Cu x 2. 1 mol Cu 1 molS 32.07 g S x x = 0.903g S 1 molS 63.55 g Cu 1 mol Cu How many grams of chlorine gas are required to react completely with 0.455 g iron to form iron(III) chloride? Answer: 0.867 g Solution What we know: g Fe Desired answer: g Cl2 First, a balanced equation is needed for the reaction between Fe and Cl2. The unbalanced equation for this combination reaction is: Fe(s) + Cl2(g) ÷ FeCl3(s) To balance this equation, place a coefficient of 2 in front of FeCl3 and a coefficient of 3 in front of Cl2 to balance chlorine atoms. This requires a coefficient of 2 in front of Fe to balance iron atoms. The balanced equation is: 2Fe(s) + 3Cl2(g) ÷ 2FeCl3(s) The solution map for this problem is: g Fe ö mol Fe ö mol Cl2 ö g Cl2 The conversion factor needed in the first step is the molar mass of Fe. From the periodic table 1mol Fe we see that the molar mass of iron is 55.84 g/mol. It is used in the form in order to 55.84 g Fe cancel units properly. S.5.32 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS 70.90 g Cl2 1mol Cl2 The conversion factor needed in the second step is the mole ratio for these two substances from 3mol Cl2 the balanced equation in the form . 2 mol Fe The conversion factor needed in the last step is the molar mass of Cl2. From the periodic table we see that the molar mass of chlorine is 35.45 g/mol so the molar mass of Cl2 is 2(35.45 g/mol) = 70.90 g/mol. It is used in the form in order to cancel units properly. Putting these together yields: 0.455 g Fe x 3. 3 mol Cl2 70.90 g Cl2 1 mol Fe x x = 0.867 g Cl2 1 mol Cl2 55.84 g Fe 2 mol Fe How many grams of each product can be formed from the decomposition of 14.0 g of sodium chlorate? 2NaClO3(s) ÷ 2NaCl(s) + 3O2(g) Answers: 7.69 g NaCl and 6.31 g O2 Solution What we know: g NaClO3; balanced equation relating NaCl, O2 and NaClO3 Desired answer: g NaCl and g O2 The solution maps for these calculations are: g NaClO3 ö mol NaClO3 ö mol NaCl ö g NaCl g NaClO3 ö mol NaClO3 ö mol O2 ö g O2 In each case, the conversion factor needed in the first step is the molar mass of NaClO3. From the periodic table we can get the molar masses of sodium, chlorine and oxygen and add them as follows. 22.99 g Na + 35.45 g Cl + 3(16.00 g) O = 106.44 g. Thus, the molar mass of NaClO3 1mol NaClO3 is 106.44 g/mol. It is used in the form in order to cancel units properly. 106.44 g NaClO3 The conversion factor needed in the second step is the mole ratio for the particular product and 3mol O 2 2 mol NaCl N2H4: and 2 mol NaClO3 2 mol NaClO3 S.5.33 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The conversion factor needed in the last step is the molar mass of the product: and 58.44 g NaCl 1mol NaCl 32.00 g O 2 1mol O 2 Putting these together yields: 14.0 g NaClO3 x 1 mol NaClO3 2 mol NaCl 58.44 g NaCl = 7.69 g NaCl x x 1 mol NaCl 106.44 g NaClO3 2 mol NaClO3 14.0 g NaClO3 x 1 mol NaClO3 3 mol O 2 32.00 g O 2 x x = 6.31g O 2 1 mol O 2 106.44 g NaClO3 2 mol NaClO3 Note that the total mass of the products equals the mass of the reactant. 4. How many grams of potassium are needed to produce 16.5 kg K2O? KNO3(s) + K(s) ÷ K2O(s) + N2(g) Answer: (unbalanced) 0.867 g Solution What we know: kg K2O; equation relating K and K2O Desired answer: gK First, a balanced equation is needed. To balance this equation, place a coefficient of 2 in front of KNO3 to balance nitrogen atoms. Then place a coefficient of 6 in front of K2O to balance oxygen atoms. Finally, a coefficient of 10 in front of K will balance potassium atoms. The balanced equation is: 2KNO3(s) + 10K(s) ÷ 6K2O(s) + N2(g) The solution map for this problem is: kg K2O ö g K2O ö mol K2O ö mol K ö g K The conversion factor needed in the first step is that between kg and g in the form 103 g . 1kg S.5.34 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The conversion factor needed in the second step is the molar mass of K2O in the form 1mol K 2 O . 94.20 g K 2 O The conversion factor needed in the third step is the mole ratio for these two substances from the 10 mol K balanced equation in the form . 6 mol K 2 O The conversion factor needed in the last step is the molar mass of K. From the periodic table we 39.10 g K see that the molar mass of potassium is 39.10 g/mol. It is used in the form in order to 1mol K cancel units properly. Putting these together yields: 16.5 kg K 2 O x 103 g K 2 O 1 mol K 2 O 10 mol K 39.10 g K x = 1.14 x104 g K x x 1 mol K 1 kg K 2 O 94.20 g K 2 O 6 mol K 2 O Theoretical Yield and Limiting Reactant 1. Which is the limiting reactant when 0.68 g magnesium reacts with 17 mmol nitrogen gas to form Mg3N2? Answer: Mg Solution What we know: g Mg; mmol N2 Desired answer: which reactant, Mg or N2, is the limiting reactant The balanced equation for this combination reaction is: 3Mg(s) + N2(g) ÷ Mg3N2(s) First, determine the limiting reactant by calculating how many moles of Mg3N2 can form from each starting amount of reactant. The solution maps for these calculations are: g Mg ö mol Mg ö mol Mg3N2 mmol N2 ö mol N2 ö mol Mg3N2 S.5.35 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS For the first calculation the conversion factor needed in the first step is the molar mass of Mg in 1mol Mg the form . 24.31g Mg The conversion factor needed for the second step is the Mg3N2/Mg mole ratio from the balanced equation. Applying these yields: 0.68 g Mg x 1mol Mg 3 N 2 1 mol Mg x = 0.0093mol Mg 3 N 2 3 mol Mg 24.31 g Mg For the second calculation the conversion factor needed in the first step is that between mmol and 10-3 mol mol in the form . 1mmol The conversion factor needed in the second step is the Mg3N2/N2 mole ratio from the balanced equation. Applying this yields: 10-3 mol N 2 1mol Mg 3 N 2 x 17 mmol N 2 x = 0.017 mol Mg 3 N 2 1 mol N 2 1 mmol N 2 Since the starting amount of Mg produces the smaller amount of Mg3N2, magnesium is the limiting reactant. 2. How many moles of AsF5 can be produced when 14 moles of arsenic react with 29 mol fluorine gas? Answer: 12 mol Solution What we know: mol As; mol F2 Desired answer: mol AsF5 The balanced equation for this combination reaction is: 2As(s) + 5F2(g) ÷ 2AsF5(g) S.5.36 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS First, determine the limiting reactant by calculating how many moles of AsF5 can form from each starting amount of reactant. The solution maps for these calculations are: mol As ö mol AsF5 mol F2 ö mol AsF5 For the first calculation the conversion factor needed is the AsF5/As mole ratio from the balanced equation. Applying this yields: 14 mol As x 2 mol AsF5 = 14 mol AsF5 2 mol As For the second calculation the conversion factor needed is the AsF5/F2 mole ratio from the balanced equation. Applying this yields: 29 mol F2 x 2 mol AsF5 = 12 mol AsF5 5 mol F2 Since the starting amount of F2 produces the smaller amount of AsF5, fluorine is the limiting reactant and the maximum moles of AsF5 is 12 moles. 3. When 26.5 g CO and 3.7 g H2 are allowed to react as shown below, CO(g) + 2H2(g) ÷ CH3OH(l) a) which is the limiting reactant? b) what is the theoretical yield in grams of CH3OH? c) how much of the reactant in excess remains? Answers: a) H2 b) 29.32 g c) 0.8 g S.5.37 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Solutions a) What we know: g CO; g H2; balanced equation relating CO, H2 and CH3OH Desired answer: which reactant, CO or H2, is the limiting reactant Determine the limiting reactant by calculating how many moles of CH3OH can form from each starting amount of reactant. The solution maps for these calculations are: g CO ö mol CO ö mol CH3OH g H2 ö mol H2 ö mol CH3OH For the first calculation the conversion factors needed are the molar mass of CO and the CH3OH/CO mole ratio. Putting these together yields: 26.5 g CO x 1mol CH 3 OH 1 mol CO x = 0.946 mol CH 3 OH 1 mol CO 28.01 g CO For the second calculation the conversion factors needed are the molar mass of H2 and the CH3OH/H2 mole ratio. Putting these together yields: 3.7 g H 2 x 1mol CH 3 OH 1 mol H 2 x = 0.918 mol CH 3 OH 2 mol H 2 2.016 g H 2 Since the starting amount of H2 produces the smaller amount of CH3OH, H2 is the limiting reactant. b) What we know: mol CH3OH produced from limiting reactant Desired answer: theoretical yield of CH3OH in grams Since the theoretical yield is the amount of product produced from the limiting reactant, the solution map for this calculation is: mol CH3OH ö g CH3OH The conversion factor needed for this calculation is the molar mass of CH3OH. S.5.38 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Applying this yields: 0.918 mol CH3 OH x c) 32.04 g CH 3 OH = 29.32 g CH 3 OH 1 mol CH 3 OH What we know: mol CH3OH produced from limiting reactant Desired answer: g CO remaining Use the theoretical yield of CH3OH to calculate how much CO was used and subtract this from the starting quantity. The solution maps for these calculations are: mol CH3OH ö mol CO ö g CO used g CO remaining = g CO initially - g CO used For the first calculation the conversion factors needed are the CO/CH3OH mole ratio and the molar mass of CO. Putting these together yields: 0.918 mol CH 3 OH x 1 mol CO 28.01g CO = 25.7 g CO used x 1 mol CO 1 mol CH 3 OH Therefore, the mass of CO that remains equals 26.5 g - 25.7 g = 0.8 g. 4. How many kilograms of the excess reactant remain when a mixture of 2.50 kg of SiO2 and 2.50 kg of carbon react? SiO2(s) + 3C(s) ÷ SiC(s) + 2CO(g) Answer: 1.00 kg Solution What we know: kg SiO2; kg C; balanced equation relating SiO2 and C Desired answer: kg of excess reactant remaining S.5.39 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS First, determine the limiting reactant by calculating how many moles of SiC can form from each starting amount of reactant. The solution maps for these calculations are: kg SiO2 ö g SiO2 ö mol SiO2 ö mol SiC kg C ö g C ö mol C ö mol SiC For the first calculation the conversion factors needed are that between kg and g, the molar mass of SiO2 and the SiC/SiO2 mole ratio from the balanced equation. Putting these together yields: 2.50 kg SiO 2 x 103 g SiO 2 1 molSiO 2 1molSiC x = 41.6 molSiC x 1 kg SiO 2 60.09 g SiO 2 1 molSiO 2 For the second calculation the conversion factors needed are that between kg and g, the molar mass of C and the SiC/C mole ratio from the balanced equation. Putting these together yields: 2.50 kg C x 103 g C 1 mol C 1molSiC = 208 molSiC x x 3 mol C 1 kg C 12.01 g C Since the starting amount of SiO2 produces the smaller amount of SiC, SiO2 is the limiting reactant and carbon is the reactant in excess. Now use the theoretical yield of SiC (41.6 mol) to calculate how much C was used and subtract this from the starting quantity. The solution maps for these calculations are: mol SiC ö mol C used ö g CO used ö kg CO used kg C remaining = kg C initially - kg C used For the first calculation the conversion factors needed are the C/SiC mole ratio, the molar mass of C and the relationship between g and kg . Putting these together yields: 41.6 molSiC x 3 mol C 12.01 g C 1kg C = 1.50 kg C used x x 3 10 g C 1 molSiC 1 mol C Therefore, the mass of C that remains equals 2.50 kg - 1.50 kg = 1.00 kg.