5.6 L of chlorine was passed through a solution weighing 300 g with a mass fraction of potassium iodide of 30%.

5.6 L of chlorine was passed through a solution weighing 300 g with a mass fraction of potassium iodide of 30%. Determine the mass of potassium chloride formed.

Given:
m solution (KI) = 300 g
ω (KI) = 30%
V (Cl2) = 5.6 L

To find:
m (KCl) -?

Decision:
1) 2KI + Cl2 => 2KCl + I2;
2) M (KI) = Mr (KI) = Ar (K) * N (K) + Ar (I) * N (I) = 39 * 1 + 127 * 1 = 166 g / mol;
M (KCl) = Mr (KCl) = Ar (K) * N (K) + Ar (Cl) * N (Cl) = 39 * 1 + 35.5 * 1 = 74.5 g / mol;
3) m (KI) = ω (KI) * m solution (KI) / 100% = 30% * 300/100% = 90 g;
4) n (KI) = m (KI) / M (KI) = 90/166 = 0.54 mol;
5) n (Cl2) = V (Cl2) / Vm = 5.6 / 22.4 = 0.25 mol;
6) n (KCl) = n (Cl2) * 2 = 0.25 * 2 = 0.5 mol;
7) m (KCl) = n (KCl) * M (KCl) = 0.5 * 74.5 = 37.25 g.

Answer: The mass of KCl is 37.25 g.