A. Correct! You successfully completed the stoichiometry problem.

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1 CLEP Chemistry - Problem Drill 09: Stoichiometry No. 1 of 10 Instructions: (1) Read the problem statement and answer choices carefully () Work the problems on paper as 1. How many grams of AgCl will precipitate out if 0.7 moles of is reacted? + AgNO 3 AgCl + Ca(NO 3 ) (A) 77 g AgCl (B) 39 g AgCl (C) 0.54 g AgCl (D) g AgCl (E) 0.07 g AgCl A. Correct! You successfully completed the stoichiometry problem. There are moles of AgCl produced for each molecule of reacted. First move from moles to moles AgCl using the balanced equation coefficient, and then move from mole AgCl to grams AgCl using the molar mass. First move from moles to moles AgCl using the balanced equation coefficient, and then move from mole AgCl to grams AgCl using the molar mass. First move from moles to moles AgCl using the balanced equation coefficient, and then move from mole AgCl to grams AgCl using the molar mass. Mole ratio: = mole AgCl Molar mass: AgCl = g AgCl 0.7 mole mole AgCl g AgCl AgCl = 77 g AgCl The correct answer is (A).

2 No. of 10. How many grams of Cl are produced if.4 g is reacted? (s) + (aq) Cl (aq) + H (g) (A) 0.47 g Cl (B) 4.7 g Cl (C) 9.4 g Cl (D) 19 g Cl (E) 0.94 g Cl Move from grams to mole of using molar mass and then move from mole to mole Cl using the balanced equation coefficients and then move from mole Cl to grams using molar mass. There is of for every of Cl. C. Correct! You successfully completed the stoichiometry calculation. Move from grams to mole of using molar mass and then move from mole to mole Cl using the balanced equation coefficients and then move from mole Cl to grams using molar mass. Move from grams to mole of using molar mass and then move from mole to mole Cl using the balanced equation coefficients and then move from mole Cl to grams using molar mass. Mole ratio: = Cl Molar mass: = 4.31 g Molar mass: Cl = 95.1 g Cl.4 g 4.31 g Cl 95.1 g Cl Cl = 9.4 g Cl Mental Math ->.4/4 = 0.1 then 0.1* 95 = closest answer is 9.4. The correct answer is (C).

3 No. 3 of How many liters of 0.50M are needed to react with 7.4 g? (aq) + (s) (aq) + H O (l) (A) 0.40 L (B) 0.0 L (C) 0.80 L (D) 0.10 L (E) 0.05 L A. Correct! You correctly completed the stoichiometry calculation. There are moles of for every of calcium hydroxide. There are moles of for every of calcium hydroxide. Move from grams calcium hydroxide to moles using molar mass. Then move from moles calcium hydroxide to moles hydrochloric acid using balanced equation ratio. Finally, move from moles hydrochloric acid to liters using the concentration. Move from grams calcium hydroxide to moles using molar mass. Then move from moles calcium hydroxide to moles hydrochloric acid using balanced equation ratio. Finally, move from moles hydrochloric acid to liters using the concentration. Mole ratio: mole consumed to make Molar mass: = g Molarity: 0.50 mole = 1 L 7.4 g / 74.1 g per mole = * = 0. mole of required 0. mole / 0.5 M ( moles/liter) = 0.4 liter 7.4 g 74.1 g mole 1 L 0.50 mole = 0.4 L The correct answer is (A).

4 No. 4 of How many liters of gas are formed if 6.0 g H CO 3 decomposes? H CO 3 (s) H O (l) + (g) (A) 6.0 L (B) 11. L (C) 44.8 L (D).4 L (E) 5.6 L The ratio of moles carbon dioxide to moles carbonic acid is 1:1. The ratio of moles carbon dioxide to moles carbonic acid is 1:1. The ratio of moles carbon dioxide to moles carbonic acid is 1:1. D. Correct! You successfully used the molar volume of a gas at STP in stoichiometry. STP = Standard Temperature and Pressure. The ratio of moles carbon dioxide to moles carbonic acid is 1:1. Mole ratio: H CO 3 decomposes to produce (g) Molar mass: H CO 3 = 6.03 g H CO 3 6 g / 6 g per mole = 1.0 mole Molarity: (g)=.4 L at STP (This is a constant that you should learn. You will see it again) 6.0 g H CO 3 H CO g H CO 3 H CO 3.4 L =.4 L The correct answer is (D).

5 No. 5 of How many grams of is produced with 15 g reacts with 15 g? (aq) + (s) (aq) + H O (l) (A) 3 g (B) g (C) 11 g (D) 46 g (E) 45 g When you have two given quantities for stoichiometry, always complete both calculations and choose the smaller answer. B. Correct! You chose the smaller answer from the two stoichiometry calculations. The ratio of hydrochloric acid to calcium chloride is :1 and the ratio of calcium hydroxide to calcium chloride is 1:1. The ratio of hydrochloric acid to calcium chloride is :1 and the ratio of calcium hydroxide to calcium chloride is 1:1. The ratio of hydrochloric acid to calcium chloride is :1 and the ratio of calcium hydroxide to calcium chloride is 1:1. Mole ratio: mole = Molar ratio: = Molar mass: = g Molar mass: = g Molar mass: = g 15 g g g =. g 15 g g = 3.3 g g mole The first thing to recognize here is that one of the reactants is probable limiting. The second thing to do it to determine which reactant is limiting. The third thing to do is to calculate the amount of that can per produced from 15 gram of the limiting reactant. The correct answer is (B).

6 No. 6 of 10 Instruction: (1) Read the problem statement and answer choices carefully () Work the problems on paper as 6. A student runs the following reaction and calculates that he should get a total of 0.75 g AgCl precipitate. After the lab is completed, he found that he only got 0.65 g AgCl. What was his percent yield? + AgNO 3 AgCl + Ca(NO 3 ) (A) % (B) 13.3 % (C) 15.4 % (D) 7.5 % (E) 86.7 % Percent yield is found by dividing the actual yield of a reaction to the theoretical yield found through stoichiometry. Percent yield is found by dividing the actual yield of a reaction to the theoretical yield found through stoichiometry. Percent yield is found by dividing the actual yield of a reaction to the theoretical yield found through stoichiometry. Percent yield is found by dividing the actual yield of a reaction to the theoretical yield found through stoichiometry. E. Correct! Percent yield is found by dividing the actual yield of a reaction to the theoretical yield found through stoichiometry. % yield = actual yield / expected yield 100% % yield = (0.65 g / 0.75 g) 100% = 86.7% The correct answer is (D).

7 No. 7 of How many grams of is needed if.0 g is reacted? (s) + (aq) Cl (aq) + H (g) (A) 6.00 g (B).67 g (C) 5.34 g (D) 3.00 g (E) 1.50 g A. Correct! You converted grams magnesium to moles magnesium and then used the mole ratio from the balanced equation to reach moles and then converted to grams hydrochloric acid. You need to convert grams to moles for magnesium and then use the molar ratio from the balanced equation to reach moles before converting back to grams. You need to convert grams to moles for magnesium and then use the molar ratio from the balanced equation to reach moles before converting back to grams. You need to convert grams to moles for magnesium and then use the molar ratio from the balanced equation to reach moles before converting back to grams. You need to convert grams to moles for magnesium and then use the molar ratio from the balanced equation to reach moles before converting back to grams. Mole ratio: = mole Molar mass: = 4.31 g Molar mass: = g.0 g 4.31 g mole g = 6.00 g Cl The correct answer is (A).

8 No. 8 of If ml is required to react with 3.7 g, what is the concentration of the? (aq) + (s) (aq) + H O (l) (A) 0.17 M (B) 0.10 M (C) 0.34 M (D) 1.0 M (E) 0.68 M First convert the moles calcium hydroxide using the molar mass and then convert to moles using the balanced equation coefficients. Finally, divide moles by the liters of the acid. You found the moles of, but the questioned asked for molarity. First convert the moles calcium hydroxide using the molar mass and then convert to moles using the balanced equation coefficients. Finally, divide moles by the liters of the acid. D. Correct! You successfully completed a solution stoichiometry calculation. First convert the moles calcium hydroxide using the molar mass and then convert to moles using the balanced equation coefficients. Finally, divide moles by the liters of the acid. Mole ratio: mole = Molar mass: = g 3.7 g mole 74.1 g = 0.10 mole Molarity = 0.10 mole / L = 1.0 M The correct answer is (D).

9 No. 9 of 10 Instruction: (1) Read the problem statement and answer choices carefully () Work the problems on paper as 9. How many grams of water are formed with 11.5 L at standard temperature and pressure? H CO 3 (s) H O (l) + (g) (A) g H O (B) 9.5 g H O (C) 4.71 g H O (D) 11.5 g H O (E) 8.60 g H O First convert liters to moles for carbon dioxide using the molar volume of a gas at STP, and then convert to moles water using the balanced equation coefficients, and finally to grams water using molar mass. B. Correct! You correctly completed a gas stoichiometry calculation using molar volume of a gas at STP. First convert liters to moles for carbon dioxide using the molar volume of a gas at STP, and then convert to moles water using the balanced equation coefficients, and finally to grams water using molar mass. First convert liters to moles for carbon dioxide using the molar volume of a gas at STP, and then convert to moles water using the balanced equation coefficients, and finally to grams water using molar mass. First convert liters to moles for carbon dioxide using the molar volume of a gas at STP, and then convert to moles water using the balanced equation coefficients, and finally to grams water using molar mass. Mole ratio: H O = Molar mass: H O = 18.0 g H O Molarity: =.4 L at STP 11.5 L.4 L H O 18.0 g H O H O = 9.5 g H O The correct answer is (B).

10 No. 10 of Stoichiometry ratios come from. (A) The masses of reactants and products in a balanced equation. (B) The number of moles of reactants and products in a balanced equation. (C) The actual mass produced in a chemical reaction. (D) The subscripts within the chemical formulas. (E) The total number of molecules. Stoichiometry is not based on masses. B. Correct! The mole ratio is the stoichimetric ratio. Stoichiometry is not based on masses. Stoichiometric ratio is not the subscripts within a molecule. Stoichiometric ratios are not based on the total number of molecules. The number of moles of reactant and products in a balanced equation determines the stoichiometric ratio. The correct answer is (B).