What volume of gas will be released when 12 g of magnesium react with 400 g of a 20% sulfuric acid solution.

| May 6, 2021 | education

Find the mass of sulfuric acid in the solution.

W = m (substance): m (solution) × 100%,

hence m (substance) = (m (solution) × w): 100%.

m (substance) = (400 g × 20%): 100% = 80 g.

Let’s find the amount of the substance H2SO4.

n = m: M.

M (H2SO4) = 98 g / mol.

n = 80 g: 98 g / mol = 0.816 mol.

Let’s find the amount of Mg substance.

M (Mg) = 24g / mol.

n = 12 g: 24 g / mol = 0.5 mol.

H2SO4 is given in excess. We solve the problem by Mg.

Let’s find the quantitative ratios of substances.

Mg + H2SO4 = MgSO4 + H2 ↑.

For 1 mole of Mg, there is 1 mole of H2.

Substances are in quantitative ratios 1: 1.

The amount of substance will be equal.

n (H2) = n (Mg) = 0.5 mol.

Let’s find the volume of CO2.

V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.

V = 0.5 mol × 22.4 L / mol = 11.2 L.

Answer: 11.2 liters.



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