1. 1.Your friend Becky is sanitizing her pool by adding Ca(ClO) 2 to the water. She wants to go swimming, but you are afraid she may have added to much Ca(ClO) 2 and will die if she swims in the pool. You know that it is dangerous to swim in a pool with more than a.05 Molar concentration of ClO - in the water. If Becky’s swimming pool holds 275 gallons of water and she added 75 grams of Ca(ClO) 2 to the water, calculate the molarity and find out if it is safe to swim in. (Hint: 1 gallon = 3.79 liters) 2. 150 grams of Hydrogen gas is reacted with 1500g O 2 gas, to form H 2 O. How many grams of water will be made? 2H (g) + O 2(g) =2H 2 O (l) 3.Balance the following equation and solve for how many grams of C0 2 are formed if 10 grams of CO are reacted with excess oxygen gas. CO+ 0 2 = CO 2 4.Given the following balanced equation, how much C 6 H 5 Br will be produced when 30g C 6 H 6 reacts with 65g of Br 2 ? C 6 H 5 + Br 2 = C 6 H 5 Br + HBr 5.Balance the following equation and then use it to find out how many grams of Na I are needed to remove 1.3 Kg of O 3 ? 6.O 3 + NA I + H 2 O = O 2 + I 2 NaOH (Hint: 1Kg = 1000g) 6.) 2C 6 H 12(l) + 5O 2(g) = 2H 2 C 6 H 8 O 4(l) + 2H 2 O (g) Use the provided balanced equation to answer the question. If the limiting reactant in this equation is C 6 H 12 and 25g of it are used, what is the theatrical yield of H 2 C 6 H 8 O 4 ? b) If after performing the experiment 33.5 g of H 2 C 6 H 8 O 4 was made, what is the percent yield? 7.) 30g of an unknown organic compound are combusted, leaving you with 733g of CO 2 and 30g of water. What is the empirical formula of the compound you just combusted? 8.) 2NaOH + CO 2 +Na 2 CO 3 + H 2 O When 1.85 mol of NaOH reacts with 1 mol Co 2 which is the limiting reactant? b.) How many moles of the excess reactant are left?

2. Answer key to the Stoich Study Guide. 1 275 Gallons H 2 0 3.79 L 1 gallon = 1045.25 L H 2 O 75 g Ca(ClO) 2 1 mole Ca(ClO) 2 141.1g 2 moles ClO 1mole Ca(ClO) 2 51.5g 1 mole ClO = 54.748g ClO M = Moles solute Liters solution = 51.5g 1042.25 L =.05M 150g H 2 1 mol H 2 2.00g H 2 =75 mol H 2 2 1500g O 2 1 mol O 2 32.0g O 2 2 mol O 1 mol 0 2 =94 moles O H 2 is the limiting reactant 75 mol H 2 2 mol H 2 O 2 mol H 2 18.0g H 2 O 1 mol H 2 O =1400g H 2 O

3. 3. 2Co + O 2 = 2CO 2 10g CO 1 mole Co 28 g 2 mole CO 2 2 mol CO 44g 1 mole CO 2 = 15.7g CO 2 4 30g C 6 H 6 1 mole C 6 H 6 78g 1 mole C 6 H 6 1 mol C 6 H 5 Br 157g 1 mole C 6 H 5 Br = 60.4g C 6 H 5 Br 65g Br 2 1 mole Br 2 160g 1 mole Br 2 1 mol C 6 H 5 Br 157g 1 mole C 6 H 5 Br = 63.8g C 6 H 5 Br C 6 H 6 is the limiting reactant, 60.4g of C 6 H 5 Br is made 5 O 3 + 2Na I + H 2 0 = O 2 + I 2 + 2NaOH 1.3 Kg O 3 1000g 1Kg 1 mole O 3 48g 2 mole NaI 1 mole O 3 = 812.5g C 6 H 5 Br 150g 1 mol NaI 6 25g C 6 H 12 1 mol C 6 H 12 84g 2 mol C 6 H 12 1 mol H 2 C 6 H 8 O 4 146g H 2 C 6 H 8 O 4 = 43.5g H 2 C 6 H 8 O 4 = theoretical Yield actual Theoretical X 100% 33.5g 43.5 X 100% b. =77% percent yield.

4. 733g CO 2 1 mol CO 2 44g =16.6 mol CO 2 30g H 2 O1 mol H 2 O 18g =16.6 mol H 2 O 16.6 mol H 2 O 1 mol H 2 O 2 mol H =33.3 mol H 16.6 mol C12g 1 mole C = 66.6% C 300 g 33.3 mol H1g 1 mole H = 11.3% C 300 g 100- (66+11.3)= 22.1% O.221 O300g =66.3g O 66.3g O1 mole O 16.6g =4.125 mol O C H O 16.6 33.3 4.13 ÷ ÷ ÷ 4 8 1 Formula is C 4 H 8 O 7

5. 8 1.85 mol NaOH 1 mole H 2 O 1 mol NaOH 18g H 2 O 1mol H 2 O =16.65g H 2 O 1 mol CO 2 1 mole H 2 O 1 mol CO 2 18g H 2 O 1mol H 2 O =18g H 2 O NaOH is the limiting reactant. 16.65g H 2 O 1 mole CO 2 1 mol H 2 O 18g H 2 O 1mol H 2 O =.925 moles CO 2 used 1 mole CO 2 -.925 moles CO 2 =.075 mol excess reactant b.