Hydrogen chloride obtained by the action of sulfuric acid on sodium chloride is oxidized with MnO2.

Hydrogen chloride obtained by the action of sulfuric acid on sodium chloride is oxidized with MnO2. The resulting chlorine displaced 25.4 g of iodine from the sodium iodide solution. Calculate the mass of sodium chloride reacted.

Given:
m (I2) = 25.4 g

To find:
m (NaCl) -?

Decision:
1) H2SO4 + NaCl => NaHSO4 + HCl ↑;
4HCl + MnO2 => MnCl2 + 2H2O + Cl2 ↑;
Cl2 + 2NaI => 2NaCl + I2;
2) M (I2) = Mr (I2) = Ar (I) * N (I) = 127 * 2 = 254 g / mol;
M (NaCl) = Mr (NaCl) = Ar (Na) * N (Na) + Ar (Cl) * N (Cl) = 23 * 1 + 35.5 * 1 = 58.5 g / mol;
3) n (I2) = m (I2) / M (I2) = 25.4 / 254 = 0.1 mol;
4) n (Cl2) = n (I2) = 0.1 mol;
5) n (HCl) = n (Cl2) * 4 = 0.1 * 4 = 0.4 mol;
6) n (NaCl) = n (HCl) = 0.4 mol;
7) m (NaCl) = n (NaCl) * M (NaCl) = 0.4 * 58.5 = 23.4 g.

Answer: The mass of NaCl is 23.4 g.