Practice Worksheet. Formulas, Equations, and Moles. Part II

Name: __KEY___________________ Chapter 3 Practice Worksheet: Formulas, Equations, and Moles: Part II Section: _________ 1) Stoichiometry: Chemical Arithmetic For each equation and starting amount and substance shown, calculate the amount of the product produced. Equation Starting amount/substance Product amount/substance S (s) + O2 (g) → SO2 (g) 2.35 moles S moles SO2 2.35 mol SO2 Si (s) + 2Cl2 (g) → SiCl4 (l) 4.1 moles Cl2 grams SiCl4 3.5 x 102 g SiCl4 3H2 (g) + N2 (g) → 2NH3 (g) 0.03445 grams H2 grams NH3 0.1952 g NH3 KCN (aq) + HCl (aq) → KCl (aq) + HCN (g) 1.09 grams HCl moles HCN 0.0299 moles HCN 2NH3 (g) + H2SO4 (aq) → (NH4)2SO4 (aq) 0.00568 grams NH3 grams (NH4)2SO4 2.21 x 10-2 g (NH4)2SO4 2NO (g) + O2 (g) → 2NO2 (g) 6.50 moles O2 moles NO2 13.0 mol NO2 2) Yields of Chemical Reactions/Limiting Reactants a. MnO2 reacts with HCl to produce MnCl2, Cl2, and H2O. Write a balanced equation for this reaction. If 0.86 moles of MnO2 and 48.2 grams of HCl react, which reagent will be used up first? How many grams of Cl2 will be produced? How many moles of the excess reagent will be left over? If 19.8 grams of Cl2 were obtained in lab, what is the percent yield? MnO2 + 4 HCl → MnCl2 + Cl2 + 2 H2O 0.86 mol MnO2 → 0.86 mol Cl2 48.2 g HCl → 1.322 mol HCl → 0.331 mol Cl2 HCl is Limiting Reagent 0.331 mol Cl2 → 23.5 g Cl2 0.331 mol Cl2 uses up 0.331 mol MnO2; 0.53 mol MnO2 left over page 1 Name: _____________________ Section: _________ b. ___ CaF2 + ___ H2SO4 → ___ CaSO4 + _2__ HF In the reaction above, you begin with 6.00 g of CaF2 and 12.592 g H2SO4. You obtain 2.86 g of HF as a product. What is the percent yield of HF? 6.00 g CaF2 → 0.154 mol HF (CaF2 is LR) 12.592 g H2SO4 → 0.257 mol HF 0.154 mol HF → 3.08 g HF (2.86 g / 3.08 g) x 100 = 92.9% yield c. ___ K3PO4 (aq) + _3__ AgNO3 (aq) → _3__ KNO3 (aq) + ___ Ag3PO4 (s) 70.5 mg 15.0 mL of 0.050 M Find the mass of precipitate formed in this reaction. 70.5 mg K3PO4 → 3.32 x 10-4 mol Ag3PO4 15.0 mL AgNO3 → 2.5 x 10-4 mol Ag3PO4 2.5 x 10-4 mol Ag3PO4 → 0.11 g Ag3PO4 3) Concentrations of Reactants in Solution: Molarity a. How many moles of solute are in the following solutions? 10.76 mL of 1.54 M HF 1.66 x 10-2 moles HF 250.0 mL of 0.99 M glucose (C6H12O6) 0.25 moles C6H12O6 50.1 mL of a 0.145 M solution of H2SO4 7.26 x 10-3 moles H2SO4 b. What is the concentration of a solution made by adding 15.5666 g of KOH in a 250.0 mL flask? 1.110 M KOH c. What mass of AgNO3 is needed to make 500.0 mL of a 1.500 M solution? 127.4 grams AgNO3 page 2 Name: _____________________ Section: _________ d. Calculate the molarity of a solution that contains 0.0345 mol NH4Cl in 400 mL of solution. 8.63 x 10-2 M e. How many grams of HNO2 are present in 35.0 mL of a 2.20 M solution of nitric acid? 3.62 grams HNO2 f. What is the concentration of 156 mL of solution made from 2.5 grams of KCl? 0.21 M KCl g. How many milliliters of 1.50 M KOH solution are needed to give 0.125 mol of KOH? 83.3 mL KOH solution h. How many grams of KMnO4 are needed to make 500 mL of solution whose concentration is 1.75 M? 138 grams KMnO4 4) Diluting Concentrated Solutions a. You have a 1.00 L bottle of 12.0 M HCl on your lab bench. How would you make a 250.0 mL solution of 1.00 M HCl? Add 20.8 mL of the 12.0 M HCl to a 250 mL volumetric flask. Add just enough water to mix the solution. Swirl the flask while adding water to the line. b. What is the concentration of a solution made by adding 15.0 mL of 18.0 M H2SO4 to 100.0 mL of water? 2.35 M H2SO4 c. A bottle of 15.0 M HCl has 27.5 mL left. What will the concentration of HCl be if water is added to the bottle to fill it to the 500.0 mL mark? 0.825 M HCl page 3 Name: _____________________ Section: _________ d. 260.0 mL of water are added to 37.8 mL of 1.66 M HCl. What is the concentration of the diluted solution? 0.211 M HCl e. How many milliliters of 3.0 M H2SO4 are required to make 450 mL of 0.10 M H2SO4? 15 mL of 3.0 M H2SO4 are needed f. How would you prepare 1.45 x 103 mL of a 1.45 M HNO3 solution using 12.0 M stock solution of HNO3? (How much water must be added to the stock solution?) Add 175 mL of stock solution to a beaker and carefully measure 1275 mL of water to add to the solution. Thoroughly mix the solution. 5) Solution Stoichiometry a. ___CaCO3 (s) + _2__ HCl (aq) → ___CaCl2 (aq) + ___H2O (l) + ___CO2 (g) What mass of CaCO3 is required to react with 25.0 mL of 0.750 M HCl? 0.01875 mol HCl → 0.009375 mol CaCO3 → 0.938 g CaCO3 b. ___ HNO3 (aq) + ___ NaOH (aq) → ___ NaNO3 (aq) + ___ H2O (l) 250.0 mL of 0.100 M HNO3 is combined with an excess of NaOH. How much H2O (in g) will be produced by this reaction? 0.0250 mol HNO3 → 0.0250 mol H2O → 0.450 g H2O c. You add 500 mL of 0.100 M AgNO3 solution to a solution containing an excess of NaCl. How many grams of AgCl precipitate will you form? (Hint: Write a net ionic equation for the precipitation of AgCl.) Ag+ (aq) + Cl- (aq) → AgCl (s) 0.0500 mol AgNO3 → 0.0500 mol AgCl → 7.16 g AgCl page 4 Name: _____________________ Section: _________ d. If you mix 0.200 L of 0.100 M Pb(NO3)2 and 0.300 L of 0.200 M NaCl, how much PbCl2 precipitate will you form? (Hint: Limiting reactant problem!) Pb(NO3)2 (aq) + 2NaCl (aq) → PbCl2 (s) + 2NaNO3 (aq) 0.0200 mol Pb(NO3)2 → 0.0200 mol PbCl2 → 5.56 g PbCl2 0.0600 mol NaCl → 0.0300 mol PbCl2 → 8.34 g PbCl2 5.56 g PbCl2 produced f. A 156.7 mL solution of 1.50 M AgNO3 is mixed with a 4.22 g of solid K3PO4. When mixed together, a new solid forms. Identify the precipitate. Determine what mass of precipitate will form. Calculate the percent yield if only 6.045 g of precipitate was formed in lab. K3PO4 → 0.019879 mol Ag3PO4 AgNO3 → 0.07835 mol Ag3PO4 0.019879 mol Ag3PO4→ 8.32 g Ag3PO4 6.045 g / 8.32 g * 100 = 72.6% yield g. Calculate the mass of the precipitate formed when 2.27 L of 0.0820 M AgNO3 are mixed with 3.06 L of 0.0664 M K2SO4. 2 AgNO3 (aq) + K2SO4 (aq) → 2 KNO3 (aq) + Ag2SO4 (s) 0.18614 mol AgNO3 → 0.09307 mol Ag2SO4 0.20318 mol K2SO4 → 0.20318 mol Ag2SO4 0.09307 mol Ag2SO4 → 29.0 g Ag2SO4 page 5 Name: _____________________ Section: _________ 6) Titration a. How many milliliters of 0.155 M HCl are needed to neutralize completely 35.0 mL of 0.101 M Ba(OH)2 solution? (Hint: Write a balanced equation first.) 2HCl (aq) + Ba(OH)2 (aq) → BaCl2 (aq) + 2H2O Find moles Ba(OH)2, convert to moles HCl, divide by concentration to find liters, convert to mL 45.6 mL HCl needed to neutralize b. How many milliliters of 2.50 M H2SO4 are needed to neutralize 75.0 g of NaOH? (Hint: Write a balanced equation first.) 2NaOH (aq) + H2SO4 (aq) → 2H2O (l) + Na2SO4 (aq) 375 mL H2SO4 c. What is the molarity of a hydrochloric acid solution if it took 30.0 mL to neutralize 48.0 mL of 0.100 M NaOH? (Hint: Write a balanced equation first.) NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l) 0.160 M HCl d. 25.0 mL of 0.050 M Ba(OH)2 neutralized 40.0 mL of nitric acid (HNO3). Determine the concentration of the acid. Ba(OH)2 (aq) + 2HNO3 (aq) → Ba(NO3)2 (aq) + 2H2O (aq) 0.063 M HNO3 (aq) page 6 Name: _____________________ Section: _________ e. What is the concentration of NaOH if it takes 25 mL of 0.75 M HCl to neutralize 16.7 mL of NaOH? (Write a balanced equation first.) NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l) 1.12 M NaOH f. How many grams of solid NaOH are required to neutralize 48.2 mL of 1.25 M H2SO4? (Write a balanced equation first.) 2NaOH (aq) + H2SO4 (aq) → 2H2O (l) + Na2SO4 (aq) find moles H2SO4, convert to moles NaOH, convert to grams NaOH 4.82 g NaOH Percent Composition and Empirical Formulas 7) What is the mass percent of each element in the following compounds? a. CaCl2 110.99 g/mol; Ca: 36.1% Cl: 63.9% b. Fe2O3 159.70 g/mol; Fe: 69.9% O: 30.1% c. C6H10S2O 162.00 g/mol; C: 44.4% H: 6.2% S: 39.5% O: 9.9% 8) Calculate the empirical formulas of compounds containing the following percentages of elements. Use the molar mass to calculate the molecular formula for that compound as well. a. 44.4% C, 6.21% H, 39.5% S, and 9.86% O; molar mass = 486.39 g/mol C6H10S2O; C18H30S6O3 b. 20.2% Al, 79.8% Cl; molar mass = 266.6 g/mol AlCl3; Al2Cl6 c. 2.1% H, 65.2% O, 32.6% S; molar mass = 195.95 g/mol H2SO4; H4S2O8 d. 19.8% C, 2.50% H, 11.6% N, 66.1% O; molar mass = 360 g/mol C2H3NO5; C6H9N3O15 page 7