Mgarcia325
13.12.2019 •
Chemistry
Ammonia is produced by the millions of tons annually for use as fertilizer. it is commonly made from n2 and h2 by the haber process. because the reaction reaches equilibrium before going completely to product, the stoichiometric amount of ammonia is not obtained. at a particular temperature and pressure, 12.17 g h2 reacts with 24.34 g n2 to form ammonia. when equilibrium is reached, 23.24 g nh3 has formed.
(a) calculate the percent yield,
(b) how many moles of n2 and h2 are present at equilibrium?
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Ответ:
The % yield is 78.15 %
At the equilibrium we have 3.418 moles of H2 and 0.869 moles of N2
Explanation:
Step 1: Data given
Mass of H2 = 12.17 grams
Mass of N2 = 24.34 grams
Mass of NH3 formed = 23.24 grams
Molar mass H2 = 2.02 g/mol
Molar mass N2 = 28 g/mol
Molar mass NH3 = 17.03 g/mol
Step 2: The balanced equation
N2 + 3H2 → 2NH3
Step 3: Calculate moles of H2
Moles H2 = Mass H2 / molar mass H2
Moles H2 = 12.17 g / 2.02 g/mol
Moles H2 = 6.025 moles
Step 4: Calculate moles of N2
Moles N2 = 24.34 g / 28.0 g/mol
Moles N2 = 0.869 moles
Step 5: Calculate limiting reactant
For 1 mole of N2 consumed, we need 3 moles of H2 to produce 2 moles of NH3
N2 is the limiting reactant. It ill completely be consumed. (0.869 moles)
H2 is in excess. There will react 3*0.869 = 2.607 moles
There will remain 6.025 - 2.607 = 3.418 moles of H2
Step 6: Calculate moles of NH3
For 1 mole of N2 consumed, we need 3 moles of H2 to produce 2 moles of NH3
For 0.869 moles of N2 we have 2*0.869 = 1.738 moles of NH3
Step 7: Calculate mass of NH3
mass NH3 = moles NH3 * molar mass NH3
Mass NH3 = 1.738 moles * 17.03 g/mol
Mass NH3 = 29.60 grams NH3 = theoretical yield
Step 8: Calculate % yield
%yield =(actual yield / theoretical yield)*100%
% yield = (23.24 grams / 29.60 grams) * 100%
% yield = 78.51 %
The % yield is 78.15 %
At the equilibrium we have 3.418 moles of H2 and 0.869 moles of N2
Ответ:
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Explanation: