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?

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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