Answer:
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.
