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kyleescott8857
18.04.2020 •
Chemistry
A student determines that according to the reaction N2(g) + 3H2(g) --> 2NH3(g), if 34g of nitrogen gas is reacted with excess hydrogen, 41g of ammonia can be produced. When the actual reaction was complete, only 38g of ammonia formed. Determine the percent yield for the reaction.
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Ответ:
The percent yield of this reaction is 92.7 %
Explanation:
Step 1: Data given
Mass of nitrogen gas (N2) = 34.0 grams
Mass of ammonia (NH3 produced = 41.0 grams
Molar mass of N2 = 28.0 g/mol
Molar mass of NH3 = 17.02 g/mol
Actual yield of ammonia = 38 grams
Step 2: The balanced equation
N2(g) + 3H2(g) → 2NH3(g)
Step 3: Calculate moles
Moles = mass / molar mass
Moles N2 = 34.0 grams / 28.0 g/mol
Moles N2 = 1.214 moles
Step 4: Calculate moles NH3
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
For 1.214 moles N2 we'll have 2* 1.214 = 2.428 moles NH3
Step 5: Calculate mass NH3
Mass NH3 = moles * molar mass
Mass NH3 = 2.428 moles * 17.02 g/mol
Mass NH3 = 41 grams
Step 6: Calculate percent yield for the reaction
Percent yield = (actuald yield / theoretical yield) * 100 %
Percent yield = (38 grams / 41 grams ) * 100 %
Percent yield = 92.7 %
The percent yield of this reaction is 92.7 %
Ответ:
The answer is D, took a test