Stoichiometry Practice Problems - Answers

Stoichiometry Practice Problems -ANSWERS 1. Problem: Determine the mass (grams) of 6.00 atoms of Americium (found in smoke detectors) 6 atoms Ac x (1 mol Ac/ 6.02 x 1023 atoms Ac) x (243 g/1 mol Ac) = 2.42 x 10-21 grams Ac 2. Empirical & Molecular Formula Problems: a) Determine the molecular formula for Freon C318: Freon C318 has a molecular mass of 200.00 g/mol and a % composition of 24.02% C by mass and remainder is F 24.02 g C x (1 mol C/12.00 g) = 2.00 mol C 2.00 mol C/2.00 mol C = 1 atom of C 75.98 g F x (1 mol F/18.9 g) = 4.00 mol F 4.00 mol F/2.00 mol C = 2 atom F for every 1 atom C Empirical formula: (CF2)n n = (200.00 g/mol) /(49.8 g/mol) = 4 Molecular formula for Freon C318 is C 4F8 One unit of (CF2) has mol. mass of 498.8 g/mol b) Determine the molecular formula for the following compound that has a molecular mass of 238.80 g/mole and a % composition of 43.64% P by mass and the reminder is O. 43.64 g P x (1 mol P/31.0 g) = 1.409 mol P 1.409 mol P / 1.409 mol P = 1 P atom 55.36 g O x (1 mol O/16.0 g) = 3.523 mol O 3.532 mol O / 1.409 mol P = 2.5 atom of O for every 1 atom of P 1 atom of P x 2 = 2 atoms of P 2.5 atoms of O x 2 = 5 atoms of O Empirical formula: (P2O5)n n = (238.80 g/mol) / 142 g/mol) ~ = to 2 Molecular formula” P4O10 One unit of (P2O5) = 142 g/mol c) Determine the molecular formula for ethanol given the following information: 5.00 grams of ethanol is burned in excess oxygen to yield 9.55 grams of carbon dioxide and 5.87 grams of water. The molecular mass of ethanol is 46.0 grams. Ethanol’s formula is composed of C, H and O atoms 5.00 g ethanol + O2 (g) → 5.87 g H20 + 9.55 g CO2 5.87 g H2O x (1 mol H2O/18.0 g) x (2 mol H atoms/1 mol H2O) x (1.00 g H/mole H) = 6.52 x 10-1 grams of H 9.55 g CO2 x (1 mol CO2/44.0 g) x (1 mol C atoms/1 mol CO2) x (12.0 g C/mole C) = 2.61 grams of C 2.61 + 0.652 = 3.62 g total mass contributed by C & H in sample of ethanol 5.00 g – 3.62 g = 1.74 g is mass contributed to O in the sample of ethanol 0.652 g H x (1 mol H/1.00 g) = 0.652 mol H 2.61 g C x (1 mol C/ 12.00 g) = 0.218 mol C 1.74 g o x (1mol O/ 16.00 g) = 0.109 mol O 0.652 mol / 0.109 mol = 5.99 or 6 H atoms for every O atom 0.210 mol / .0.109 mol = 2.00 or 2 C atoms for every O atom 0.109 mol / 0.109 mol = 1 atom of O Empirical formula for ethanol is (C2H6O)n The mass of one unit of (C2H6O) is 46.0 g/mole n =(46.0 g/mol) / (46.0 g/mol) = 1 Molecular Formula for ethanol is C2H6O 3. Problems: Balance the following reactions 2 NO + 1 O2 → 2 NO2 2 H2O2 → 2 H2O + 1 O2 2 S + 3 O2 → 2 SO3 4 NH3 + 3 Cl2 → 3 NH4Cl + 1 NCl3 1 C2H5OH + 3 O2 → 3 H2O + 2 CO2 1 C6H12O6 + 6 O2 → 6 H2O + 6 CO2 1 SbCl3 + 1 H2O → 1 SbOCl + 2 HCl 3 Mg(ClO)2 + 2 AlPO4 → 1 Mg3(PO4)2 + 2 Al(ClO)3 2 Fe + 3 S → 1 Fe2S3 2 ZnS + 3 O2 → 2 ZnO + 2 SO2 1 N2O4 + 2 N2H4 → 4 H2O + 3 N2 2 C8H18 + 25 O2 → 18 H2O + 16 CO2 1 CH4 + 3Cl2 → 3 HCl + 1 CHCl3 4. Stoichiometry Problems a) The chief component of glass is silicon dioxide. To dissolve silicon dioxide, hydrofluoric acid is used and the following products are made, solid silicon tetrafluoride and water vapor. Determine how many grams of silicon tetrafluoride would be produced if 63.4 grams of hydrofluoric acid reacted with excess silicon dioxide? Determine the actual grams of HF produced a 75% yield was observed SiO2 (s) + 4 HF (g) → SiF4 (s) + 2 H2O (g) 63.4 g HF x (1 mol HF/20.0 g) x (1 mol SiF4/ 4 mol HF) x (104.1 g SiF4/1 mol SiF4) = 82.4 g SiF4 This is your theoretical yield Let x = grams actual produced: (x/ 82.4 g) = 0.75 x = 61.8 g SiF4 made b) 18.10 grams of ammonia gas was reacted with 90. 40 grams of solid cupric oxide to produce nitrogen gas + copper metal + water vapor. Determine the % yield of this reaction if 8.90 grams of nitrogen gas was produced. Balanced reaction: 2 NH3 + 3 CuO → N2 + 3 Cu + 3 H2O 18.10 g NH3 x (1 mol NH3/17.0 g) x (1 mol N2/2 mol NH3) x (28.0 g/1 mol N2) = 14.83 g N2 (to be made) 90.40 g CuO x (1 mol CuO/79.6 g) x (1 mol N2/3 mol CuO) x (28.0 g/1 mol N2) = 10.60 g N2 (to be made) 10.60 g < 14.83 g of N2 so the limiting reactant is CuO and the theoretical yield is 10.60 g N2 Percent yield: (8.90 g/10.60 g) x 100 = 83.96% or 84.0 % Determine the amount in grams of excess reagent from this reaction. NH3 is the reactant in excess. 90.40 g CuO x (1 mol CuO/79.6 g) x (2 mol NH3/3 mol CuO) x (17.0 g/1 mol NH3) = 12.87 g NH3 (used in reaction) 18.10 g- 12.87 g = 5.23 g of NH3 remains or is in excess. c) When solid iron (III) oxide reacts with heated carbon monoxide, iron metal and carbon dioxide is produced. Determine the actual mass of iron metal produced, in grams, given that the theoretical yield for the iron metal was determined to be 0.600 moles of iron metal and the % yield of the overall reaction was 85.1% 0.600 mol Fe x (55.85 g/ 1 mol Fe) = 33.5 g (theoretical yield) Let x = actual yield: (x/33.5 g) = 0.851 x = 28.5 g is actual yield of Fe d) Cisplatin is an antitumor drug. Given the following information, determine the % yield when 35.50 grams of cisplatin is produced. 65.50 g K2PtCl4 + 5.10 g NH3 (g) → KCl (aq) + Pt(NH3)2Cl2 (s) Balanced reaction: K2PtCl4 + 2 NH3 (g) → 2 KCl (aq) + 1 Pt(NH3)2Cl2 (s) 65.50 g K2PtCl4 x (1 mol K2PtCl4/413 g) x (1 mol Pt(NH3)2Cl2 / 1 mol K2PtCl4) x (299.99 g/ mol Pt(NH3)2Cl2) = 47.42 g Pt(NH3)2Cl2 (theoretical) 5.10 g NH3 x (1 mol NH3/17.0 g) x (1 mol Pt(NH3)2Cl2 /1 mol NH3) x (299.99 g/ mol Pt(NH3)2Cl2) = 44.82 g Pt(NH3)2Cl2 (theoretical) 47.42 g > 44.82 g Pt(NH3)2Cl2 so NH3 is the limiting reactant and 44.82 grams of Pt(NH3)2Cl2 is the theoretical yield. Percent yield: (35.50 g/44.82 g) x 100 = 79.20% e) A 0.900 gram sample of magnesium metal is placed in a flask containing 0.250 grams of nitrogen gas. The flask is heated and 0.785 grams of magnesium nitride is produced. The chemical equation for this reaction is given below: Magnesium metal + nitrogen gas → magnesium nitride (g) i. Write a balanced chemical equation for this reaction, using chemical formulas. 3 Mg (s) + N2 (g) → Mg3N2 (g) Ii. Calculate the percent yield for this chemical reaction. 0.900 g Mg x (1 mol Mg /24 .0 g) x (1 mol Mg3N2/3 mol Mg) x (100.0 g/1 mol Mg3N2) = = 1.25 g Mg3N2 (theoretical) 0.250 g N2 x (1 mol N2/28.0 g) x (1 mol Mg3N2/1mol N2) x (100.0 g/1 mol Mg3N2) = 0.893 g Mg3N2 (theoretical) 0.893 g < 1.25 g of Mg3N2 so N2 is the limiting reactant and 0.893 grams Mg3N2 is the theoretical yield Percent yield: (0.785 g/0.893 g) x 100 = 87.9% iii. Determine in micrograms which reactant, if any was in excess? Mg is the reactant in excess: 0.250 g N2 x (1 mol N2/28.0 g) x (3 mol Mg/1mol N2) x (24.0 g/1 mol Mg) = 0.643 g Mg was used 0.900 g – 0.643 g = 0.257 g of Mg was in excess f) The thermite reaction was once used by the railroads to weld rails. The thermite reaction is: iron (III) oxide (s) + aluminum (s) + heat → aluminum oxide (s) + iron (l) i. Using chemical formulas, balance the above redox reaction. Fe2O3 (s) + 2 Al (s) → Al2O3 (s) + 2 Fe (s) ii. Determine the mass of liquid iron produced when 25.0 grams of aluminum metal and 85.0 grams of iron (III) oxide solid are mixed and heated together. 25.0 g Al x (1 mol Al/26.9 g) x (2 mol Fe/2 mol Al) x (55.85 g/1 mol Fe) = 51.9 g Fe (theoretical) 85.0 g Fe2O3 x (1 mol Fe2O3/159.7 g) x (2 mol Fe/1 mol Fe2O3) x (55.85 g/ 1 mole Fe) = 59.5 g Fe (theoretical) 51.9 g < 59.5 g Fe so the limiting reactant is Al and the theoretical yield is 51.9 g Fe iii. Determine the percent yield of the reaction if 45.8 grams of liquid iron is produced. Percent yield: (45.8 g/51.9 g) x 100 = 88.3%