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How many milliliters of sodium metal, with a density of 0.97 g/mL, would be needed to produce 43.6 grams of sodium hydroxide in the single replacement reaction below? Be sure to show the work that you did to solve this problem.

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Answer is: 25.84  milliliters of sodium metal.

Balanced chemical reaction: 2Na + 2H₂O → 2NaOH + H₂.

d(Na) = 0.97 g/mL; density of sodim.

m(NaOH) = 43.6 g; mass of sodium hydroxide.

n(NaOH) = m(NaOH) ÷ M(NaOH).

n(NaOH) = 43.6 g ÷ 40 g/mol.

n(NaOH) =1.09 mol; amount of sodium hydroxide.

From chemical reaction: n(NaOH) : n(Na) = 2 : 2 (1: 1).

n(Na) = 1.09 mol.

m(Na) = 1.09 mol · 23 g/mol.

m(Na) = 25.07 g; mass of sodium.

V(Na) = m(Na) ÷ d(Na).

V(Na) = 25.07 g ÷ 0.97 g/mL.

V(Na) = 25.84 mL.

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