Chlorine with a volume of 5.6 L was passed through a solution containing 60 g of sodium iodide.

Chlorine with a volume of 5.6 L was passed through a solution containing 60 g of sodium iodide. Determine the mass of the gas formed?

Given:
m (NaI) = 60 g
V (Cl2) = 5.6 L
Vm = 22.4 l / mol

To find:
m (gas) -?

Decision:
1) 2NaI + Cl2 => 2NaCl + I2;
2) M (NaI) = Mr (NaI) = Ar (Na) * N (Na) + Ar (I) * N (I) = 23 * 1 + 127 * 1 = 150 g / mol;
M (I2) = Mr (I2) = Ar (I) * N (I) = 127 * 2 = 254 g / mol;
3) n (NaI) = m (NaI) / M (NaI) = 60/150 = 0.4 mol;
4) n (Cl2) = V (Cl2) / Vm = 5.6 / 22.4 = 0.25 mol;
5) n (I2) = n (NaI) / 2 = 0.4 / 2 = 0.2 mol;
6) m (I2) = n (I2) * M (I2) = 0.2 * 254 = 50.8 g.

Answer: The mass of I2 is 50.8 g.