lecture17%28chapt9%29

lecture17%28chapt9%29 - Chapter 9 Calculations from...

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1 Chapter 9 Calculations from Chemical Equations % Yield 15.00 g aluminum sulfide and 10.00 g water react until the limiting reagent is used up. Here is the balanced equation for the reaction: Al 2 S 3 + 6 H 2 O -----------> 2Al(OH) 3 + 3 H 2 S (A) Which is the limiting reagent? (B) What is the maximum mass of H 2 S which can be formed from these reagents? (C) How much excess reagent remains after the reaction is complete?
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2 15.00 g aluminum sulfide and 10.00 g water react until the limiting reagent is used up. Here is the balanced equation for the reaction: Al 2 S 3 + 6 H 2 O -----------> 2Al(OH) 3 + 3 H 2 S (A) Which is the limiting reagent? 15.00 g Al 2 S 3 X 150g Al 2 S 3 1 mol Al 2 S 3 = 10.80 g H 2 O 1 mol Al 2 S 3 6 mol H 2 O X 18 g H 2 O 1 mol H 2 O X 15.00 g aluminum sulfide and 10.00 g water react until the limiting reagent is used up. Here is the balanced equation for the reaction: Al 2 S 3 + 6 H 2 O -----------> 2Al(OH) 3 + 3 H 2 S (A) Which is the limiting reagent? (B) What is the maximum mass of H 2 S which can be formed from these reagents? 10.00 g H 2 O X 18 g H 2 O 1 mol H 2 O = 9.444 g H 2 S 6 mol H 2 O 3 mol H 2 S X 34 g H 2 S 1 mol H 2 S X
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3 15.00 g aluminum sulfide and 10.00 g water react until the limiting reagent is used up. Here is the balanced equation for the reaction: Al 2 S 3 + 6 H 2 O -----------> 2Al(OH) 3 + 3 H 2 S (A) Which is the limiting reagent? (B) What is the maximum mass of H 2 S which can be formed from these reagents? (C) How much excess reagent remains after the reaction is complete? 10.00 g H 2 O X 18 g H 2 O 1 mol H 2 O 6 mol H 2 O X 150g Al 2 S 3 1 mol Al 2 S 3 1 mol Al 2 S 3 X = 13.89 g Al 2 S 3 15.00-13.89 = 1.11g Al 2 S 3
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4 The typical calculation in a starting class assumes that there is only one path for the reactants. Food is not always converted to energy. If you eat a cookie, some of it could end up stored as "fat" Reactions rarely produce the predicted amount of product from the masses of reactants in the reaction. C(s) + O 2 (g) --- > CO 2 (g) Normally we expect a 1 mol yield of carbon dioxide for every mol of carbon burned. 12 g C X 1 mol C 12 g C 1 mol CO 2 1 mol C 44 g CO 2 1 mol CO 2 = 44 g CO 2 X X 12 g 44 g Grams A to mols A mols A to mols B mols B to grams B
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5 C(s) + O 2 (g) --- > CO 2 (g) the amount you will get is more like 34 grams of CO 2 . a competing reaction
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lecture17%28chapt9%29 - Chapter 9 Calculations from...

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