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PERCENT YIELD

PERCENT YIELD. Frequently the amount of a product made during a chemical reaction is less that that predicted by mass – mass stoichiometry. Percent yield is calculated by dividing the amount actually produced by the theoretical amount as determined by mass – mass stoichiometry and then

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PERCENT YIELD

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  1. PERCENT YIELD Frequently the amount of a product made during a chemical reaction is less that that predicted by mass – mass stoichiometry. Percent yield is calculated by dividing the amount actually produced by the theoretical amount as determined by mass – mass stoichiometry and then multiplying by 100.

  2. Percent Yield Sample Problem: What is the percent yield if 17.5 grams of barium hydroxide (Ba(OH)2 when reacting with an excess of hydrochloric acid actually produces 2.93 grams of water? Note: put the grams of given substance over the barium hydroxide and the grams of water actually produced over the water. Since you want to determine the theoretical amount of water put a “? “over it as well. ? 17.5 g 2.92 g actual HCl + Ba(OH)2 BaCl2 + H2O 2 2 Step 1: Balance and solve a mass – mass problem for the theoretical amount of water which can be produced. 17.5 g Ba(OH)2 1 mol Ba(OH)2 2 mol H2O 18 g H2O ____________________________________________________ = 3.68 g H2O 171 g Ba(OH)2 1 mol Ba(OH)2 1 mol H2O Step 2: Divide the actual by the theoretical and multiply by 100. 2.92 gx 100 = 3.68g 79.6 % yield

  3. Practice Problems Try to work the next problems on the paper first. Then use the power point to check your work.

  4. Practice Problem 1: What is the percent yield if the decomposition of 10.9 grams of sodium bicarbonate (NaHCO3) actually yields 1.71 grams of carbon dioxide (CO2)? ? 10.9 g 1.71 g NaHCO3 Na2CO3 + CO2 + H2O 2 Solve and then check your answer below. 10.9 g NaHCO3 1 mol NaHCO3 1 mol CO2 44.0 g CO2 ___________________________________________________ = 2.85 g CO2 84.0 g NaHCO3 2 mol NaHCO3 1 mol CO2 = 60.0 % yield 1.71 gx 100 2.85 g

  5. Practice Problem 2: What is the percent yield if 13.4 g of sulfuric acid (H2SO4) reacts with an excess of sodium nitrate (NaNO3) to actually produce 14.8 g of nitric acid (HNO3)? ? 13.4 g 14.8 g actual H2SO4 + NaNO3 Na2SO4 + HNO3 2 2 Solve and then check your answer below. _________________________________________________ 13.4 g H2SO4 1 mol H2SO4 2 mol HNO3 63.0 g HNO3 = 17.2 g HNO3 98.1 g H2SO4 1 mol H2SO4 1 mol HNO3 86.0 % yield 14.8 gx 100 = 17.2 g

  6. Practice Problem 3: What is the percent yield if 9.16 g of silver nitrate reacts with an excess of iron (Fe) to actually produce 5.77 grams of silver? ? 9.16 g 5.77 g actual Fe + AgNO3 Ag + Fe(NO3)3 3 3 Solve and then check your answer below. 9.16 g AgNO3 ____________ ______________________________ 1 mol AgNO3 3 mol Ag 108 g Ag 5.82 g Ag = 170 g AgNO3 3 mol AgNO3 1 mol Ag 5.77 g x 100 5.82 g = 99.1 % yield

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