Question

the electrolysis of aqueous potassium iodide
stoichiometry
the mass of iodine produced during the reaction is
3.95 g i₂.

calculate the dry volume of hydrogen gas produced at
stp during the electrolysis.

2i⁻(aq) + 2h₂o(l) → h₂(g) + 2oh⁻(aq) + i₂(s)

what volume of h₂ is formed during the reaction at
stp?
0.697 l
3.95 l
0.349 l

Explanation:

Step1: Calculate moles of \( I_2 \)

Molar mass of \( I_2 \) is \( 2\times126.90 = 253.80 \, \text{g/mol} \).
Moles of \( I_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{3.95 \, \text{g}}{253.80 \, \text{g/mol}} \approx 0.01556 \, \text{mol} \).

Step2: Relate moles of \( I_2 \) to \( H_2 \)

From the reaction: \( 1 \, \text{mol} \, I_2 \) produces \( 1 \, \text{mol} \, H_2 \).
Thus, moles of \( H_2 = 0.01556 \, \text{mol} \).

Step3: Volume of \( H_2 \) at STP

At STP, 1 mol of gas occupies \( 22.4 \, \text{L} \).
Volume of \( H_2 = 0.01556 \, \text{mol} \times 22.4 \, \text{L/mol} \approx 0.349 \, \text{L} \).

Answer:

0.349 L