Calculate the volume of hydrogen in the interaction of H2so4 (9.8g) and Mg (6g)

1. Determine the amount of sulfuric acid and magnesium.

n = m: M,

M (H2SO4) = 1 × 2 + 32 + 16 × 4 = 98g / mol

M (Mg) = 24g / mol

n (H2SO4) = 9.8: 98 g / mol = 0.1 mol.

n (Mg) = 6: 24 g / mol = 0.25 mol.

2. Let’s compose the equation of the reaction between magnesium and sulfuric acid.

Mg + H2SO4 = MgSO4 + H2.

Mg and H2SO4 react in quantitative ratios of 1: 1, we will find the volume of hydrogen by deficiency (for that substance, the amount of which is less, for sulfuric acid, n (H2SO4) = 0.1 mol).

Let’s find the quantitative ratios of sulfuric acid and hydrogen.

1 mol H2SO4 – 1 mol H2, are in a 1: 1 ratio, so the amount of substance will be the same.

n (H2SO4) = n (H2) = 0.1 mol.

3.Let’s find the volume of hydrogen by the formula:

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 = 22.4 L / mol × 0.1 mol = 2.24 L.

Answer: V = 2.24 liters.