Presentation on theme: "12.3 Limiting Reagent and Percent Yield"— Presentation transcript:
1 12.3 Limiting Reagent and Percent Yield By:Michelle Alice Ganian
2 Limiting ReagentIn 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 + 3CuWhat is the limiting reagent?Step 1: Change grams to moles to find the limiting reagent54g 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 productLimiting reagent = 2 mole CuSO42 mole CuSO4 x 3 mole Cu/3 mole CuSO4 x 63.5g Cu/1 mole Cu = 127g Cu
5 Percent YieldActual yield - the amount of product that actually forms when a reaction is carried out in a labTheoretical yield- the maximum amount of product that can be formed from given amounts of reactantsPercent 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 + CO2Given mass CaCO3 = 24.8gMolar mass CaCO3 = gMolar mass CaO = 56.1gTheoretical 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% = 94.2Hint: the larger number always goes as the denominator