Determining Percent Yield First, you need a balanced chemical equation - reagent

Question

Determining Percent Yield First, you need a balanced chemical equation - reagents and products (solvents and catalysts are not part of chemical equations; solvents & catalysts are NOT reagents). Determine the number of moles of each reagent used. Determine the number of reaction equivalents of each reagent used coefficient (the coefficient is the number in front of each chemical in the balanced chemical equation). whichever reagent is present in the least quantity (lowest # equivalents, that is) is termed the LIMITING REAGENT. This quantity of the limiting reagent n - this number of equivalents-represents the maximum amount of product (the maximum-equivalents") that can be formed given 100% reaction. 1. 2. 3, # moles/ reaction 4, 5. Determine the theoretical yield of product (moles) = maximum # "equivalents" (n) * 6. Determine the theoretical yield of product (grams) theoretical yield (moles) MW of 7. % Yield-100% * actual yield of product obtained (grams) / theoretical yield product coefficient of product. product. (grams) Examples 2H2 + O2-> 2H20 Pt catalyst 1. If 2.00 g hydrogen gas is reacted with 12.0 g oxygen gas in the presence of 0.40 g Pt catalyst to produce 5.00 g water. What is (a) the limiting reagent (show calculations), (b) the theoretical yield of water (in moles), and (c) the percent yield? 2. If 2.00 g hydrogen gas is reacted with 28.0 g oxygen gas in the presence of 0.40 g Pt catalyst to produce 5.00 g water. What is (a) the limiting reagent (show calculations), (b) the theoretical yield of water (in moles), and (c) the percent yield? Give your answers to the above two examples on a separate sheet of paper. Due beginning of class (week 3). 5 pts each. Please work alone (no collaboration).

Explanation / Answer

Q1)

The balanced reaction is

2H2(g) + O2(g) ------- 2H2O(g)

Number of moles of H2 = Mass/molar mass = 2/2 = 1 mol

Number of moles of O2 = Mass/molar mass = 12/32 = 0.375 mol

1 mole of O2 reacts with 2 moles of H2, hence 0.375 moles of O2 will react with 0.750 moles of H2

(a) limiting reagent = O2

(b) Number of moles of H2O produced = 2 * Number of moles of O2 = 2 * 0.375 = 0.75 moles

Theoritical Yield of water = number of moles * molar mass = 0.75 mol * 18 gm/mol = 13.5 grams

(c) Percent Yield = Actual Yield/Theoritical Yield * 100 = 5/13.5 * 100 = 37.037%

Q2)

The balanced reaction is

2H2(g) + O2(g) ------- 2H2O(g)

Number of moles of H2 = Mass/molar mass = 2/2 = 1 mol

Number of moles of O2 = Mass/molar mass = 28/32 = 0.875 mol

1 mole of O2 reacts with 2 moles of H2, hence 0.875 moles of O2 will react with 1.750 moles of H2

(a) limiting reagent = H2

(b) Number of moles of H2O produced = Number of moles of H2 = 1 mol

Theoritical Yield of water = number of moles * molar mass = 1 mol * 18 gm/mol = 18 grams

(c) Percent Yield = Actual Yield/Theoritical Yield * 100 = 5/18 * 100 = 27.77%

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