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Limiting Reactant and Percent Yield

Limiting Reactant and Percent Yield. The reactant that is consumed first and limits the formation of products. Limiting Reactant. When 2 substances are allowed to react one of them usually reacts completely while the other is not all used up because it is present in excess

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Limiting Reactant and Percent Yield

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  1. Limiting Reactant and Percent Yield The reactant that is consumed first and limits the formation of products

  2. Limiting Reactant • When 2 substances are allowed to react one of them usually reacts completely while the other is not all used up because it is present in excess • The reactant that is completely used up in the reaction is the limiting reactant. It determines the amount of products formed. The other reactant that is not completely consumed in the reaction is the excess reactant

  3. Percent Yield In chemical calculations, it is always assumed that the reaction goes to completion. However in reality, this seldom happens. The expected amount of product is not usually obtained percent yield = actual yield x 100% theoretical yield Actual yield -> amount of product formed from the actual chemical reaction and is usually less than the theoretical yield Theoretical yield the maximum amount of products which could be produced by the complete reaction of the limiting reactant

  4. Example1 • 2Al + 3I2 ----> 2AlI3 Determine the limiting reactant if one starts with • 1.20mol Al and 2.40 mol I2 • 1.20gAl and 2.40g I2 Solution • Since the given is already in moles, we use those numbers directly. To find the limiting reactant: take the moles of each substance and divide it by the coefficient of the balanced equation. Al : 1.20/2 = 0.6 I2 : 2.40/3 = 0.80 the lowest number indicates the limiting reactant • Since we have grams, we must first convert to moles. Then multiply it by the molar ratio of the reactant to product. 1.20g Al x 1molAl = 0.044molAl x 2molAlI3 = 0.044 AlI3 26.98gAl 2molAl 2.40gI2 x 1molI2= 0.0095molI2 x 2molAlI2 = 0.0033 molAlI3 253.8gI2 3mol I2 I2 is the limiting reactant since it produces less amount of the product

  5. Example In an experiment, 5.00g aluminum is heated with 25gS to form aluminum sulfide. The equation for the reaction is 2Al(s) + 3S(s) ------> Al2S3(s) • How many grams of aluminum sulfide will be formed? • How many grams of excess reactant will remain. Solution Step1: Identify the limiting reactant by calculating the amount of product formed from each of the given amounts of reactants. 5g Al x 1mol Al = 0.185 mol Al 26.98gAl From the balanced equation, 2 mol Al forms 1 molAl2S3, therefore mole Al2S3: 0.185mol Al x 1mol Al2S3 = 0.0925mol Al2S3 2mol Al 25gS x 1mol S = 0.780molS 32.07gS From the balance equation 3moles S forms 1mole Al2S3 therefore Mole Al2S3 : 0.780S x 1molAl2S3 = 0.260mol Al2S3 3molS 5gAl produces less amount of Al2S3 therefore, it is the limiting reactant

  6. Calculate the mass of Al2S3 0.0925Al2S3mol x 150.17g Al2S3 = 13.9gAl2S3 1molAl2S3 b. Find the excess amount by determining the actual amount of S that reacted with the limiting reactant, Al From the balanced equation, 2 moles Al reacts with 3 moles S; therefore mole S : 0.185mol Al x 3mol S = 0.278mol S 2mol Al Since the available S is 0.780mol then the amount of S that has not reacted is 0.780mol-0.278mol = 0.502mol S Calculating the mass excess 0.502mol S X 32.07gS = 16.1g 1mol S

  7. Sample Problem • Wine is produced by the fermentation of fruit sugar, fructose, to alcohol. The chemical reaction is C6H12O6 ---> 2C2H6O + 2CO2 If 938g of fructose was used in the preparation of wine, what is the percent yield if, after the fermentation,327g ethanol was produced? Find: % yield Solution: % yield = actual Yield x 100% theoretical yield • Determine the theoretical yield Convert mass C6H12O6 to moles using molar mass and calculate the maximum number of moles of C2H6O which could be produced using the molar ratio in the balanced equation 938gC6H12O6 x 1mole C6H12O6 x 2molC2H6O = 10.4molC2H6O 180gC6H12O6 1molC6H12O6 then convert 10.4 molC2H6O using its molar mass (46.07g/mol) 10.4mol C2H6O x 46.07g = 479g 1mol C2H6O 2. Calculate the percent yield = Act. Yield x 100% = 327g x100% = 68.3% Theo.Yield 479g

  8. Practice Problems 1.Consider the reaction: Se + 3BrF5 ----> SeF6 + 3 BrF3 , if 0.270mole Se reacts with BrF and 83.4gBrF3 is formed, what is the percent yield of the reaction? 2.The reaction: 4NH3 +5 O2 ---> 4NO + 6H2O is one of the steps in the commercial processes for converting ammonia(NH3) to nitric acid(HNO3). In a certain experiment, 4.50gNH3 is reacted with 7.50gO2 a) Which is the limiting reactant? b) How many grams of H2O are formed assuming 100%yield? c) How many grams of excess reactant remain after the reaction? Answers: • % yield = 75.1% 2. a) O2 b) 5.06gH2O c) 1.41gNH3

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