What weight of iodine is liberated from a solution of potassium iodine

1.	What weight of iodine is liberated from a solution of potassium iodine when 1 litre of Cl2 gas at 10°C and 750 mm pressure is passed through it?
Answer» The reaction is 2KI + Cl₂ → 2KCl + I₂ 22.4 L at STP 2 × 127 = 254 g (V₁) volume of Cl₂ gas = 1 L, V₂ = volume of gas at STP =? (P₁) Pressure = \(\frac{750}{760}\) atm, P₂ = atm Temperature (T₁) = 10°C + 273 = 283 K, T₂ = 273 K \(\therefore\) V₂ = \(\frac{P_1V_1}{T_1}\) x \(\frac{T_2}{P_2}\) = \(\frac{750\times1\times173}{760\times283\times1}\) = 0.952 L \(\therefore\) Volume of Cl₂ (g) passed at STP = 0.952 L Now 22.4 of Cl₂ produces at STP = 254 g of I₂ 0.952 L of Cl₂ at STP will produce = \(\frac{254}{22.4}\) x 0.952 g of l₂ = 10.795 g of l₂.

What weight of iodine is liberated from a solution of potassium iodine when 1 litre of Cl2 gas at 10°C and 750 mm pressure is passed through it?

Answer»

The reaction is 2KI + Cl₂ → 2KCl + I₂

22.4 L at STP 2 × 127 = 254 g

(V₁) volume of Cl₂ gas = 1 L, V₂ = volume of gas at STP =?

(P₁) Pressure = \(\frac{750}{760}\) atm, P₂ = atm

Temperature (T₁) = 10°C + 273 = 283 K,

T₂ = 273 K

\(\therefore\) V₂ = \(\frac{P_1V_1}{T_1}\) x \(\frac{T_2}{P_2}\)

= \(\frac{750\times1\times173}{760\times283\times1}\) = 0.952 L

\(\therefore\) Volume of Cl₂ (g) passed at STP = 0.952 L

Now 22.4 of Cl₂ produces at STP = 254 g of I₂

0.952 L of Cl₂ at STP will produce

= \(\frac{254}{22.4}\) x 0.952 g of l₂

= 10.795 g of l₂.