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Question #22091
When 9.0 g of hydrogen gas, H2, reacts with oxygen gas, O2, 73.0 g of water is produced.

a.) What is the theoretical yield of water?

b.) What is the percentage yield?
1
2013-01-17T10:33:25-0500
Theequation for this reaction is
2H2 + O2 = 2H2O;
Having a mass of hydrogen we can find the quantity substance
n(H2) = m(H2)/M(H2);
M(H2) = 1 *2 = 2 g/mol
n(H2) = 9 g / 2 g/mol = 4.5 moles;
The amount of substance of hydrogen with the amount of substance of water as
a 2 to 2 or 1 to 1.
so n(H2O) = 4.5 moles
Theoretical mass of water will be
n(H2O) = m(H2O)/M(H2O);
M(H2O) = 1 *2 + 16 = 18 g/mol
m (H2O) 4.5 moles * 18 g/mol = 81 g.
so the theoretical yield of water is 81 g.
The percentage yield will be
m(H2O)practical / m(H2O)theoretical = 73.0 g / 81.0 g = 0.9012 or 90.12 %

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