B. Incorrect! There are 2 moles of AgCl produced for each molecule of CaCl 2 reacted. It is not 1:1 ratio.

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1 AP Chemistry - Problem Drill 19: Stoichiometry No. 1 of 10 Instructions: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as 1. A balanced chemical equation gives the mole ratios of reactants and products in a chemical reaction. Use such a mole ratio to connect reactant to product via moleto-mole conversion. How many grams of AgCl will precipitate out if 0.27 moles of is reacted? (aq) + 2 AgNO 3 (aq) 2 AgCl (s) + Ca(NO 3 ) 2 (aq) (A) 77 g AgCl (B) 39 g AgCl (C) 0.54 g AgCl (D) g AgCl A. Correct! Good job! Apply the mole ratio of the balanced equation to calculate the moles of the product AgCl using the moles of the reactant, and then convert the product AgCl moles to grams. Set up the dimensional analysis to make all the conversions in one setting. There are 2 moles of AgCl produced for each molecule of reacted. It is not 1:1 ratio. Apply the mole ratio of the balanced equation to calculate the moles of the product AgCl using the moles of the reactant, and then convert the product AgCl moles to grams. Set up the dimensional analysis to make all the conversions in one setting. Apply the mole ratio of the balanced equation to calculate the moles of the product AgCl using the moles of the reactant, and then convert the product AgCl moles to grams. Set up the dimensional analysis to make all the conversions in one setting. Mole ratio: e = 2 moles AgCl (stoichiometric ratio of the reaction) Molar mass: e AgCl = g AgCl 0.27 mol 2 mol AgCl g AgCl AgCl = 77 g AgCl The correct answer is (A).

2 No. 2 of Stoichiometry is the chemical calculation of quantitative relationships of reactants and products in a balanced equation. The key in stoichiometric calculations is to convert everything to moles first so the mole ratio of the balanced equation can be used to calculate the unknown. How many grams of Cl 2 are produced if 2.4 g is reacted? (s) + 2 (aq) Cl 2 (aq) + H 2 (g) (A) 0.47 g Cl 2 (B) 4.7 g Cl 2 (C) 9.4 g Cl 2 (D) 19 g Cl 2 Convert grams to moles of using molar mass of. Next convert moles to moles Cl 2 using the balanced equation coefficients (stoichiometric ratio). Finally convert the moles Cl 2 to grams using molar mass of Cl 2 for the answer. According to the reaction mole ratio, there is e of for every e of Cl 2. Recheck your calculation. C. Correct! Good job! Convert grams to moles of using molar mass of. Next convert moles of to moles of Cl 2 using the balanced equation coefficients (stoichiometric ratio). Finally convert the moles of Cl 2 to grams using molar mass of Cl 2 for the answer. Convert grams to moles of using molar mass of. Next convert moles to moles Cl 2 using the balanced equation coefficients (stoichiometric ratio). Finally convert the moles Cl 2 to grams using molar mass of Cl 2 for the answer. Mole ratio: e = e Cl 2 Molar mass: e = g Molar mass: e Cl 2 = g Cl g Cl g Cl g Cl 2 = 9.4 g MCl 2 Mental Math 2.4/24 = 0.1 then 0.1* 95 = closest answer is 9.4. The correct answer is (C).

3 No. 3 of Molarity is the molar concentration of a solution. It is calculated by taking the moles of solute and dividing by the liters of solution, i.e. moles of solute per liter of solution. How many liters of 0.50M are needed to react with 7.4 g? 2 (aq) + (s) (aq) + 2 H 2 O (aq) (A) 0.40 L (B) 0.20 L (C) 0.80 L (D) 0.10 L A. Correct! Good job! When comes to the stoichiometric calculation, always convert whatever is given to the moles first and then use the stoichiometric ratio to get to the other species (also in moles). Finally convert the moles to grams or whatever asked. There are 2 moles of for every e of calcium hydroxide. This is not 1:1 ratio. There are 2 moles of for every e of calcium hydroxide. Recheck the calculation and be sure the correct mole ratio is used. Convert the grams to moles first before using the mole ratio of the balanced reaction to calculate the moles of. Finally convert moles to volume of using the molarity given. Mole ratio: 2 moles of consumed to make e Molar mass: e = g Molarity: 0.50 moles = 1 L 7.4 g / 74.1 g per mole = 0.es 0.es * 2 = 0.2 moles of required 0.2 moles / 0.5 M ( moles/liter) = 0.4 liter 7.4 g 74.1 g 2 mol 1 L 0.50 mol = 0.4 L The correct answer is (A).

4 No. 4 of Molar volume is the volume occupied by a mole of a substance. At STP, the molar volume of a gas is 22.4 L/mol. How many liters of gas are formed if 62.0 g H 2 CO 3 decomposes? H 2 CO 3 (aq) H 2 O (aq) + (g) (A) 62.0 L (B) 11.2 L (C) 44.8 L (D) 22.4 L The ratio of moles carbon dioxide to moles carbonic acid is 1:1. Convert the grams to moles first. Apply the molar volume of a gas at STP in stoichiometry. The ratio of moles carbon dioxide to moles carbonic acid is 1:1. Convert the grams to moles first. Apply the molar volume of a gas at STP in stoichiometry. The ratio of moles carbon dioxide to moles carbonic acid is 1:1. Convert the grams to moles first. Apply the molar volume of a gas at STP in stoichiometry. D. Correct! Good job! Convert the grams to moles first. Apply the molar volume of a gas at STP in stoichiometry. STP = Standard Temperature and Pressure. Mole ratio: e H 2 CO 3 decomposes to produce e (g) Molar mass: e H 2 CO 3 = g H 2 CO 3 62 g / 62 g per mole = 1.0 mole Molarity: e (g) = 22.4 L at STP (Standard Temperature & Pressure) 62.0 g H 2 CO 3 H 2 CO g H 2 CO 3 H 2 CO L = 22.4 L The correct answer is (D).

5 No. 5 of Limiting reactant is the reactant that limits the amount of product formed. One way to find the limiting reactant is to calculate and compare the amount of product each reactant produced. The reactant that produces a lesser amount of product is the limiting reactant. How many grams of is produced with 15 g reacts with 15 g? 2 (aq) + (s) (aq) + 2 H 2 O (aq) (A) 15 g (B) 22 g (C) 11 g (D) 46 g First determine the limiting reactant from two reactants and (the masses) provided. Use the limiting reactant and mole ratio to calculate the moles of produced. B. Correct! Good job! First determine the limiting reactant from two reactants and (the masses) provided. Use the limiting reactant and mole ratio to calculate the amount of produced. First determine the limiting reactant from two reactants and (the masses) provided. Use the limiting reactant and mole ratio to calculate the amount of produced. First determine the limiting reactant from two reactants and (the masses) provided. Use the limiting reactant and mole ratio to calculate the amount of produced. Mole ratio: 2 moles = e Molar ratio: e = e Molar mass: e = g Molar mass: e = g Molar mass: e = g 15 g g g = 22.2 g - limiting reactant. 15 g g 2 mol g CaCl = 23.3 g 2 - excessive reactant. (1) Determine which is the limiting reactant; (2) Use the mole ratio to obtain the moles of product produced by this limiting reactant; (3) Convert moles to grams. The correct answer is (B).

6 No. 6 of 10 Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as 6. Reactions rarely produce the predicted amount of products from the masses of reactants. A percent yield of a reaction is to measure the efficiency of a reaction. 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? (aq) + 2 AgNO 3 (aq) 2 AgCl (s) + Ca(NO 3 ) 2 (aq) (A) 101 % (B) 13 % (C) 15.4 % (D) 87 % Percent yield is found by dividing the actual yield of a reaction to the theoretical yield found through stoichiometry. The percent yield should be less than 100%. 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. D. Correct! Good job! Percent yield is found by dividing the actual yield of a reaction to the theoretical yield found through stoichiometry. Notice the significant figures for both yields. % yield = (actual yield / expected yield) 100% % yield = (0.65 g / 0.75 g) 100% = 87% The correct answer is (D).

7 No. 7 of The first step of stoichiometry is to convert all given data into moles, since the stoichiometric ratio in a balanced equation is the mole ratio. The second step is to use the mole ratio to calculate the moles of the target species (usually the product). The last step is to convert the moles back to grams if asked. Let s practice. How many grams of is needed if 2.00 g is reacted? (s) + 2 (aq) Cl 2 (aq) + H 2 (g) (A) 6.00 g (B) 2.67 g (C) 5.34 g (D) 3.00 g A. Correct! Good job! Converted grams of magnesium to moles of magnesium and then used the mole ratio from the balanced equation to reach moles of 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. Mole ratio: e = 2 moles Molar mass: e = g Molar mass: e = g 2.00 g g 2 mol g = 6.00 g There are three significant figures in the given data and the same applies to the final answer. The correct answer is (A).

8 No. 8 of The typical stoichiometric calculation takes three steps, after a reaction equation has been balanced. (1) Convert units of a given substance to moles; (2) Apply the mole ratio to calculate the moles of a substance produced. (3) Convert the moles of the target substance to desired units. Let s put this process in action. If ml is required to react with 3.7 g, what is the concentration of the? 2 (aq) + (s) (aq) + 2 H 2 O (aq) (A) 0.17 M (B) 0.10 M (C) 0.34 M (D) 1.0 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 question asks 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! Good job! Follow the three-step process and be aware of the significant figures at the end. Mole ratio: 2 mole = e Molar mass: e = g 3.7 g 74.1 g 2 mol = 0.10 moles Molarity = (0.10 moles ) / (0.100 L ) = 1.0 M The correct answer is (D).

9 No. 9 of 10 Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as 9. Carbonic acid in water is also called carbonated water. can be treated as an ideal gas with a molar volume of 22.4 L/moles. How many grams of water are formed with 11.5 liters of at standard temperature and pressure? H 2 CO 3 (aq) H 2 O (aq) + (g) (A) g H 2 O (B) 9.25 g H 2 O (C) 4.71 g H 2 O (D) 11.5 g H 2 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 of water using molar mass. B. Correct! Good job! You correctly completed a gas stoichiometry calculation using molar volume of a gas at STP. Apply the mole ratio for the conversion between the two products (H 2 O and ). 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 of 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 of water using molar mass. Mole ratio: e H 2 O = e Molar mass: e H 2 O = g H 2 O Molarity: e = 22.4 L at STP 11.5 L 22.4 L H 2 O g H 2 O H 2 O = 9.25 g H 2 O The correct answer is (B).

10 No. 10 of Stoichiometry calculation is based on the stoichiometric ratios of a reaction. Stoichiometric 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. Stoichiometry is not based on masses. The mole ratio is the stoichimetric ratio. The subscripts in a formula are in mole ratio and the coefficients in an equation are also in mole ratio. For stoichiometry, convert everything to moles first before using the stoichiometric ratios for the calculations. B. Correct! Good job! The mole ratio is the stoichimetric ratio. The subscripts in a formula are in mole ratio and the coefficients in an equation are also in mole ratio. For stoichiometry, convert everything to moles first before using the stoichiometric ratios for the calculations. Stoichiometry is not based on masses. The mole ratio is the stoichimetric ratio. The subscripts in a formula are in mole ratio and the coefficients in an equation are also in mole ratio. For stoichiometry, convert everything to moles first before using the stoichiometric ratios for the calculations. Stoichiometric ratio is not the subscripts within a molecule. The mole ratio is the stoichimetric ratio. The subscripts in a formula are in mole ratio and the coefficients in an equation are also in mole ratio. For stoichiometry, convert everything to moles first before using the stoichiometric ratios for the calculations. The number of moles of reactant and products in a balanced equation determines the stoichiometric ratio. The mole ratio is the stoichimetric ratio. The subscripts in a formula are in mole ratio and the coefficients in an equation are also in mole ratio. For stoichiometry, convert everything to moles first before using the stoichiometric ratios for the calculations. The correct answer is (B).