# Sodium nitrite 6.9 was added by heating to 110 g of ammonium chloride solution with a mass fraction of 10%.

Sodium nitrite 6.9 was added by heating to 110 g of ammonium chloride solution with a mass fraction of 10%. What volume of nitrogen will be released in this case and what is the mass fraction of sodium chloride in the resulting solution?

Find the mass of NH4Cl in solution.

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

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

m (NH4Cl) = (110 g × 10%): 100% = 11 g.

Let’s find the amount of the substance NH4Cl.

М (NH4Cl) = 58.5 g / mol.

n = m: M.

n (NH4Cl) = 11 g: 53.5 g / mol = 0.2056 mol.

m (H2O) = 110 g – 11 g = 99 g.

Let’s find the amount of the substance NaNO2.

M (NaNO2) = 69 g / mol.

n (NaNO2) = 6.9 g: 69 g / mol = 0.1 mol.

NaNO2 + NH4Cl = N2 + NaCl + 2H2O.

n (NaNO2) = n (N2) = 0.1 mol.

V (N2) = 22.4 L / mol × 0.1 mol = 2.24 L.

n (NaNO2) = n (NaCl) = 0.1 mol.

M (NaCl) = 58.5 g / mol.

m (NaCl) = 0.1 mol × 58.5 = 5.85 g.

n (NaNO2) = n (NH4Cl) = 0.1 mol.

n (NH4Cl) = 0.2056 – 0.1 = 0.1056 mol.

m (NH4Cl) = 0.1056 × 53.5 = 5.65 g.

m (solution) = m (H2O) + m (NH4Cl) unreacted + m (NaCl) = 99 g + 5.65 + 5.85 = 110.5 g.

W (NaCl) = (5.85: 110.5 g) × 100% = 5.29%.