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?

Expert's answer

Theequation for this reaction is

2H_{2} + O_{2} = 2H_{2}O;

Having a mass of hydrogen we can find the quantity substance

n(H_{2}) = m(H_{2})/M(H_{2});

M(H_{2}) = 1 *2 = 2 g/mol

n(H_{2}) = 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(H_{2}O) = 4.5 moles

Theoretical mass of water will be

n(H_{2}O) = m(H_{2}O)/M(H_{2}O);

M(H_{2}O) = 1 *2 + 16 = 18 g/mol

m (H_{2}O) 4.5 moles * 18 g/mol = 81 g.

so the theoretical yield of water is 81 g.

The percentage yield will be

m(H_{2}O)practical / m(H_{2}O)theoretical = 73.0 g / 81.0 g = 0.9012 or 90.12 %

2H

Having a mass of hydrogen we can find the quantity substance

n(H

M(H

n(H

The amount of substance of hydrogen with the amount of substance of water as

a 2 to 2 or 1 to 1.

so n(H

Theoretical mass of water will be

n(H

M(H

m (H

so the theoretical yield of water is 81 g.

The percentage yield will be

m(H

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