How many grams of sodium metal must be introduced to water to produce 3.3 grams of hydrogen gas?

The mass of sodium metal that must be introduced to water to produce 3.3 grams of hydrogen gas, H₂, is approximately 18.82 grams of sodium metal

Explanation:

The given mass of hydrogen gas produced = 3.3 grams

The molar mass of hydrogen gas, H₂ = 2.016 g/mol

The number of moles of hydrogen gas in 3.3 grams of H₂, 'n', is given as follows;

n = Mass/(Molar mass)

n = 3.3 g/(2.016 g/mol) = 1.63690476 moles of H₂

The reaction of sodium and water can be written as follows;

2Na + 2H₂O → 2NaOH + H₂ (g)

2 moles of sodium produces 1 mole of hydrogen gas, H₂

Therefore;

1.63690476/2 moles of sodium will produce 1.63690476 moles of hydrogen gas, H₂

The molar mass of sodium, Na ≈ 22.989 g/mol

The mass of 1.63690476/2 moles of sodium, 'm', is given as follows;

m = 1.63690476/2 moles × 22.989 g/mol ≈ 18.8154018 grams ≈ 18.82 grams

The mass of sodium that will produce 3.3 grams of hydrogen, m ≈ 18.82 grams of sodium metal.

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