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12.3 Limiting Reagent and Percent Yield

12.3 Limiting Reagent and Percent Yield. By: Michelle Alice Ganian. Limiting Reagent. In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that is formed.

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12.3 Limiting Reagent and Percent Yield

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  1. 12.3 Limiting Reagent and Percent Yield By: Michelle Alice Ganian

  2. Limiting Reagent In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that is formed. (In terms of ingredients, the smallest amount you have of the ingredients will be the limiting reagent, because it is the substance that will limit the amount of the recipe.) Limiting Reagent-the reagent that determines the amount of product that can be formed by the reaction. When the limiting reagent is used up, the chemical reaction is over. Excess reagent-the reaction that is not completely used up in an equation.

  3. Finding the Limiting Reagent • How many grams of metallic copper can be obtained when 54g Al react with 319g of CuSO4? • 2Al + 3CuSO4 1Al2(SO4)3 + 3Cu • What is the limiting reagent? • Step 1: Change grams to moles to find the limiting reagent • 54g Al x 1 mole Al/27g Al = 2 mole Al (you have 2Al in the equation) • 319g CuSO4 x I mole CuSO4/159.5g CuSO4 = 2 mole CuSO4

  4. Using the Limiting Reagent How many grams of metallic copper can be obtained when 54g Al react with 319g of CuSO4? Step 2: Continue the problem by using the founded limiting reagent to solve for the quantity of the product Limiting reagent = 2 mole CuSO4 2 mole CuSO4 x 3 mole Cu/3 mole CuSO4 x 63.5g Cu/1 mole Cu = 127g Cu

  5. Percent Yield Actual yield - the amount of product that actually forms when a reaction is carried out in a lab Theoretical yield- the maximum amount of product that can be formed from given amounts of reactants Percent yield – the ratio of the actual yield to the theoretical yield expressed as a percent.

  6. Calculating Theoretical Yield • What is the theoretical yield of CaO if 24.8g of CaCO3 is heated? • CaCO3 CaO3 + CO2 Given mass CaCO3 = 24.8g Molar mass CaCO3 = 100.1 g Molar mass CaO = 56.1g Theoretical yield is unknown. 24.8g CaCO3 x 1 mole CaCO3/100.1g CaCO3 x 1 mole CaO/1 mole CaCO3 x 56.1g CaO/ 1 mole CaO = 13.9g CaO

  7. Calculating for Percent Yield What is the percent yield if 13.1g CaO is actually produced when 24.8g CaCO3 is heated? (You don’t need 24.8 g CaCO3) Percent yield = actual yield/theoretical yield x 100% Percent yield =13.1g CaO/13.9g CaO x 100% = 94.2 Hint: the larger number always goes as the denominator

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