Through 200 g of potassium iodide solution, 100 ml of a gas mixture containing chlorine was passed.

Through 200 g of potassium iodide solution, 100 ml of a gas mixture containing chlorine was passed. In this case, 0.508 g of crystalline iodide was released. What is the mass fraction (in%) of potassium iodide in the analyzed solution? How much chlorine was contained in the gas mixture? (Chlorine and potassium iodide reacted completely)

Given:
m solution (KI) = 200 g
V mixture (Cl2) = 100 ml
m (I2) = 0.508 g

Find:
ω (KI) -?
V (Cl2) -?

Solution:
1) 2KI + Cl2 => 2KCl + I2;
2) n (I2) = m (I2) / M (I2) = 0.508 / 254 = 0.002 mol;
3) n (KI) = n (I2) * 2 = 0.002 * 2 = 0.004 mol;
4) m (KI) = n (KI) * M (KI) = 0.004 * 166 = 0.664 g;
5) ω (KI) = m (KI) * 100% / m solution (KI) = 0.664 * 100% / 200 = 0.332%;
6) n (Cl2) = n (I2) = 0.002 mol;
7) V (Cl2) = n (Cl2) * Vm = 0.002 * 22.4 = 0.0448 l.

Answer: Mass fraction of KI is 0.332%; Cl2 volume – 0.0448 l.