What weight of iodine is liberated from a solution, of potassium iodide when 1 liter of Cl2 gas at 10° C and 750 mm pressure is passed through it?
The chemical equation is 2KI + Cl2 → 2KCl + I2 22.4 L 2 × 127 at STP = 254 g (V1) volume of Cl2 gas=1 L V2 = vol. of gas at STP =? (P1) Pressure = 750760atm P2 = 1 atm (T1) Temperature = 10°C + 273 = 283 K T2 = 273 K ∴ V2 = P1V1T−1×T2P2=750×1×273760×283×1 = 0.952 L ∴ Vol of Cl2 g passed at STP = 0.952 L Now 22.4 L of Cl2 produces at STP = 254 g of I2 0.952 L of Cl2 at STP produces = 25422.4 × 0.952 g of I2 = 10.78 g of I2.
