Practice Problems: Balancing & Stoichiometry KEY CHEM 1A I suggest that you complete these practice problems in pencil since you may need to erase and change coefficients as you balance the chemical equations. Note: coefficients of 1 can be omitted, and are only shown here for clarity Balance the following equations (show your check). 1. 2 SO 2 + 1 O 2 2 SO 3 2. 4 Al + 3 MnO 2 3 Mn + 2 Al 2 O 3 3. 1 Bi 2 S 3 + 6 HCl 2 BiCl 3 + 3 H 2 S 4. 2 PbO 2 2 PbO + 1 O 2 5. 3 H 2 SO 4 + 2 Al(OH) 3 6 H 2 O + 1 Al 2 (SO 4 ) 3 6. 1 C 3 H 8 + 5 O 2 3 CO 2 + 4 H 2 O Write formula equations from the following word equations, then balance them (show your check). 7. phosphoric acid + calcium hydroxide calcium phosphate + water 2 H 3 PO 4 + 3 Ca(OH) 2 1 Ca 3 (PO 4 ) 2 + 6 H 2 O 8. zinc carbonate + hydrochloric acid zinc chloride + water + carbon dioxide 1 ZnCO 3 + 2 HCl 1 ZnCl 2 + 1 H 2 O + 1 CO 2 9. silver nitrate + aluminum chloride silver chloride + aluminum nitrate 3 AgNO 3 + 1 AlCl 3 3 AgCl + 1 Al(NO 3 ) 3 S 2 O 6 Pb 2 O 4 Al 4 Mn 3 O 6 C 3 H 8 O 10 10. silver oxide silver + oxygen 2 Ag 2 O 4 Ag + 1 O 2 H 12 Bi 2 S 3 H 6 Cl 6 S 3 O 18 Al 2 Ag 4 O 2 H 12 P 2 O 14 Ca 3 Ag 3 N 3 O 9 Al 1 Cl 3 Zn 1 C 1 O 3 H 2 Cl 2
11. How many molecules of CO 2 are in 12.0 g CO 2? 12.0 g CO 2 1 mol CO 2 6.022 x 10 23 molecules CO 2 44.01 g CO 2 1 mol CO 2 = 1.64 x 10 23 molecules CO 2 12. What is the average mass (in grams) of 1 atom of gold? 1 atom Au 1 mol Au 6.022 x 10 23 atoms Au 197.0 g Au 1 mol Au = 3.271 x 10 22 g Au 13. In the following reaction, 23.2 grams of butane (C 4 H 10 ) and 93.7 grams of oxygen (O 2 ) are available to react, and 69.2 g CO 2 product are obtained. 2 C 4 H 10 + 13 O 2 8 CO 2 + 10 H 2 O 23.2 g 93.7 g 69.2 g C 8 H 20 O 26 a) balance the equation b) determine the limiting reactant c) calculate the theoretical yield of CO 2 (in grams), and d) calculate the % yield of CO 2. Way #1: 23.2 g C 4 H 10 L.R. 93.7 g O 2 Way #2: 23.2 g C 4 H 10 1 mol C 4 H 10 58.120 g C 4 H 10 1 mol O 2 32.00 g O 2 1 mol C 4 H 10 58.120 g C 4 H 10 theoretical yield CO 2 8 mol CO 2 44.01 g CO 2 = 70.271 g CO 2 forms 2 mol C 4 H 10 1 mol CO 2 less product 8 mol CO 2 44.01 g CO 2 = 79.303 g CO 2 13 mol O 2 1 mol CO 2 13 mol O 2 32.00 g O 2 = 83.028 g O 2 only need 2 mol C 4 H 10 1 mol O 2 have 93.7 g O 2 1 mol C 4 H 10 23.2 g C 4 H 10 L.R. 58.120 g C 4 H 10 (actually have) 8 mol CO 2 44.01 g CO 2 = 70.271 g CO 2 2 mol C 4 H 10 1 mol CO 2 theoretical yield CO 2 actual yield CO 2 % yield CO 2 = theoretical yield CO 2 (100%) = 69.2 g CO 2 70.271 g CO 2 (100%) = 98.476% yield CO 2 b) C 4 H 10 c) 70.3 g CO 2 d) 98.5 % yield CO 2
14. Imitation pineapple flavoring is 62.04% C, 10.41% H, and 27.55% O by mass, and has a molar mass of 116.156 g/mol. What are the empirical and molecular formulas of this food additive? Assume a 100.00 g sample of the compound: 1 mol C C 62.04 g C = 5.165695254 mol C = 3.000 12.01 g C H 1 mol H 10.41 g H = 10.32738095 mol H 1.721875 = 5.998 6 1.008 g H moles O 27.55 g O 1 mol O 16.00 g O = 1.721875 mol O = 1.000 Empirical formula = C 3 H 6 O mass compound 116.156 g/mol Scaling Factor = = = 2 mass C 3 H 6 O 58.078 g/mol C 3 H 6 O x 2 C 6 H 12 O 2 empirical formula multiply subscripts by scaling factor molecular formula Empirical Formula: C 3 H 6 O Molecular Formula: C 6 H 12 O 2
15. When 10.029 g of an unknown compound containing only C, H, and O are combusted, 21.015 g of CO 2 and 4.299 g of H 2 O are produced. What is the empirical formula of the compound? If the molar mass of the compound is 126.1 g/mol, what is the molecular formula? 126.1 g/mol C x H y O z + O 2 CO 2 + H 2 O 10.029 g 21.015 g 4.299 g 21.015 g CO 2 1 mol CO 2 44.01 g CO 2 1 mol C 1 mol CO 2 = 0.4775051 mol C 12.01 g C 1 mol C = 5.734836 g C 4.299 g H 2 O 1 mol H 2 O 18.016 g H 2 O 2 mol H 1 mol H 2 O = 0.4772424 mol H 1.008 g H 1 mol H = 0.481060 g H 10.029 g total 5.734836 g C 0.481060 g H 3.813103 g O 0.4775051 mol C = 2.004 0.4772424 mol H 0.2383190 mol = 2.003 3.813103 g O 1 mol O 16.00 g O = 0.2383190 mol O = 1.000 Empirical formula = C 2 H 2 O mass compound 126.1 g/mol Scaling Factor = = = 3 mass C 2 H 2 O 42.036 g/mol C 2 H 2 O x 3 C 6 H 6 O 3 empirical formula multiply subscripts by scaling factor molecular formula Empirical Formula: C 2 H 2 O Molecular Formula: C 6 H 6 O 3
16. Describe in detail how you would make 500.0 ml of a 1.750 M CuCl 2 solution from solid CuCl 2. Include the mass of the solute needed, the procedure (in order), and any specific laboratory equipment you would need. 1) Determine how many grams of solid CuCl 2 are required soln. 1.750 mol CuCl 2 134.456 g CuCl 2 500.0 m = 117.649 g CuCl 2 = 117.6 g CuCl 2 soln. 1 1 mol CuCl 2 2) Measure out 117.6 g CuCl 2 on a balance 3) Completely dissolve this mass of CuCl 2 in a portion of the solvent water 4) Dilute the solution with enough water to bring the meniscus up to the etched calibration line on the neck of a 500.0 ml volumetric flask 5) Transfer the solution to an appropriately labeled storage container 6) Stick a smiley-face sicker on your forehead 17. To make 250.0 ml of a 2.25M HCl solution, what volume of 12.1 M HCl stock solution should be diluted to 250.0 ml in a volumetric flask? M 1 V 1 = M 2 V 2 (12.1 M) V 1 = (2.25M) (250.0 ml) V 1 = 46.5 ml 18. If 88.4 ml of a 1.50 M KMnO 4 stock solution is diluted to 100.0 ml, and a 10.00 ml aliquot of the resulting diluted solution is further diluted to 250.0 ml, then what is the concentration of KMnO 4 in the final solution? M 1 V 1 = M 2 V 2 (1.50 M) (88.4 ml) = M 2 (100.0 ml) M 2 = 1.326 M M 2 V 2 = M 3 V 3 (1.326 M) (10.00 ml) = M 3 (250.0 ml) M 3 = 0.05304 M = 0.0530 M KMnO 4 19. What is the final concentration of Cl if 128.5 ml of 0.950 M CaCl 2 is mixed with 245.8 ml of 0.150 M NaCl? 128.5 ml 0.950 mol CaCl 2 2 mol Cl 1 mol CaCl 2 = 0.24415 mol Cl 245.8 ml 0.150 mol NaCl 1 mol Cl 1 mol NaCl = 0.03687 mol Cl 0.244 15 mol Cl (from CaCl 2 soln.) + 0.036 87 mol Cl (from NaCl soln.) 0.281 02 mol Cl (total in mixed soln.) (128.5 ml + 245.8 ml) = 0.3743 L mixed soln. Molarity = mol solute L solution 0.28102 mol Cl = = 0.7507881379 M Cl = 0.751 M Cl 0.3743
20. H + and OH neutralize each other according to the following equation: H + + OH H 2 O What is the final concentration of H + if 275.0 ml of 0.750 M HBr is mixed with 125.0 ml of 0.925 M KOH? 275.0 ml 0.750 mol HBr 1 mol H + 1 mol HBr = 0.20625 mol H + 125.0 ml 0.925 mol KOH 1 mol OH 1 mol KOH = 0.115625 mol OH 1 mol H + 0.115625 mol OH = 0.115625 mol H + 1 mol OH 0.206 25 mol H + (in original HBr soln.) 0.115 625 mol H + (used to neut. OH - ) 0.090 625 mol H + (left in mixed soln.) (275.0 ml + 125.0 ml) = 0.4000 L mixed soln. Molarity = mol solute L solution 0.090625 mol H + = = 0.2265625 M H + = 0.23 M H + 0.4000