Note: coefficients of 1 can be omitted, and are only shown here for clarity. S 2 O 6 charge 0 Check: Al 4 Mn 3 O 6. charge 0. Pb 2 O 4.

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1 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) SO O 2 2 SO Al + 3 MnO 2 3 Mn + 2 Al 2 O Bi 2 S HCl 2 BiCl H 2 S 4. 2 PbO 2 2 PbO + 1 O H 2 SO Al(OH) 3 6 H 2 O + 1 Al 2 (SO 4 ) C 3 H O 2 3 CO 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 Ca(OH) 2 1 Ca 3 (PO 4 ) H 2 O 8. zinc carbonate + hydrochloric acid zinc chloride + water + carbon dioxide 1 ZnCO HCl 1 ZnCl H 2 O + 1 CO 2 9. silver nitrate + aluminum chloride silver chloride + aluminum nitrate 3 AgNO 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 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

2 11. How many molecules of CO 2 are in 12.0 g CO 2? 12.0 g CO 2 1 mol CO x molecules CO g CO 2 1 mol CO 2 = 1.64 x molecules CO What is the average mass (in grams) of 1 atom of gold? 1 atom Au 1 mol Au x atoms Au g Au 1 mol Au = x 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 O 2 8 CO 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 g O 2 Way #2: 23.2 g C 4 H 10 1 mol C 4 H g C 4 H 10 1 mol O g O 2 1 mol C 4 H g C 4 H 10 theoretical yield CO 2 8 mol CO g CO 2 = g CO 2 forms 2 mol C 4 H 10 1 mol CO 2 less product 8 mol CO g CO 2 = g CO 2 13 mol O 2 1 mol CO 2 13 mol O g O 2 = 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 g C 4 H 10 L.R g C 4 H 10 (actually have) 8 mol CO g CO 2 = 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 g CO 2 (100%) = % yield CO 2 b) C 4 H 10 c) 70.3 g CO 2 d) 98.5 % yield CO 2

3 14. Imitation pineapple flavoring is 62.04% C, 10.41% H, and 27.55% O by mass, and has a molar mass of g/mol. What are the empirical and molecular formulas of this food additive? Assume a g sample of the compound: 1 mol C C g C = mol C = g C H 1 mol H g H = mol H = g H moles O g O 1 mol O g O = mol O = Empirical formula = C 3 H 6 O mass compound g/mol Scaling Factor = = = 2 mass C 3 H 6 O 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

4 15. When g of an unknown compound containing only C, H, and O are combusted, g of CO 2 and g of H 2 O are produced. What is the empirical formula of the compound? If the molar mass of the compound is g/mol, what is the molecular formula? g/mol C x H y O z + O 2 CO 2 + H 2 O g g g g CO 2 1 mol CO g CO 2 1 mol C 1 mol CO 2 = mol C g C 1 mol C = g C g H 2 O 1 mol H 2 O g H 2 O 2 mol H 1 mol H 2 O = mol H g H 1 mol H = g H g total g C g H g O mol C = mol H mol = g O 1 mol O g O = mol O = Empirical formula = C 2 H 2 O mass compound g/mol Scaling Factor = = = 3 mass C 2 H 2 O 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

5 16. Describe in detail how you would make ml of a 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 mol CuCl g CuCl m = g CuCl 2 = g CuCl 2 soln. 1 1 mol CuCl 2 2) Measure out 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 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 ml of a 2.25M HCl solution, what volume of 12.1 M HCl stock solution should be diluted to 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 ml, and a ml aliquot of the resulting diluted solution is further diluted to 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 = M M 2 V 2 = M 3 V 3 (1.326 M) (10.00 ml) = M 3 (250.0 ml) M 3 = M = M KMnO What is the final concentration of Cl if ml of M CaCl 2 is mixed with ml of M NaCl? ml mol CaCl 2 2 mol Cl 1 mol CaCl 2 = mol Cl ml mol NaCl 1 mol Cl 1 mol NaCl = mol Cl mol Cl (from CaCl 2 soln.) mol Cl (from NaCl soln.) mol Cl (total in mixed soln.) (128.5 ml ml) = L mixed soln. Molarity = mol solute L solution mol Cl = = M Cl = M Cl

6 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 ml of M HBr is mixed with ml of M KOH? ml mol HBr 1 mol H + 1 mol HBr = mol H ml mol KOH 1 mol OH 1 mol KOH = mol OH 1 mol H mol OH = mol H + 1 mol OH mol H + (in original HBr soln.) mol H + (used to neut. OH - ) mol H + (left in mixed soln.) (275.0 ml ml) = L mixed soln. Molarity = mol solute L solution mol H + = = M H + = 0.23 M H