# A solution of nitric acid with a mass of 25.2 g and a mass fraction of 10% was added to an excess

**A solution of nitric acid with a mass of 25.2 g and a mass fraction of 10% was added to an excess of sodium carbonate. Calculate the amount of gas evolved.**

Find the mass of HNO3 in solution.

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

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

m (HNO3) = (25.2 g × 10%): 100% = 2.52 g.

Let’s find the amount of substance HNO3 by the formula:

n = m: M.

M (HNO3) = 63 g / mol.

n = 2.52 g: 63 g / mol = 0.04 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

2HNO3 + Na2CO3 = 2NaNO3 + CO2 + H2O.

According to the reaction equation, there is 1 mol of CO2 for 2 mol of HNO3. Substances are in quantitative ratios of 2: 1.

n (CO2) = ½ n (HNO3) = 0.04: 2 = 0.02 mol.

Let’s find the volume of CO2.

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

V = 0.02 mol × 22.4 L / mol = 0.448 L.

Answer: 0.448 l.