Question

From the following reaction sequence:
$C{l}_2+2KOH\to KCl+KClO+H_{2}O$
$3KClO\to 2KCl+KCl{O}_3$
$4KCl{O}_3\to 3KCl{O}_4+KCl$
Calculate the mass of chlorine that reacts with an excess of $KOH$ needed to produce $100\text{g}$ of $KCl{O}_4$.

• A. $195.05\text{g}$
• B. $205.05\text{g}$
• C. $280.66\text{g}$
• D. $328.78\text{g}$

Answer 1

The correct option is B $205.05\text{g}$

Given reactions:
$4KCl{O}_3\to 3KCl{O}_4+KCl$
$12KClO\to 8KCl+4KCl{O}_3$
$12C{l}_2+24KOH\to 12KClO+12H_{2}O+12KCl$

Overall reaction: $12C{l}_2+24KOH\to 3KCl{O}_4+21KCl+12H_{2}O$

Molar mass of $KCl{O}_4=138.5\text{g}$
3 mol of $KCl{O}_4$ are formed from 12 mol of $C{l}_2$.
$3\times 138.5\text{g}$ of $KCl{O}_4$ is formed from $12\times 71\text{g}$ of $C{l}_2$.
So,$100\text{g}$ of $KCl{O}_4$ will be formed =$\frac{12\times 71\times 100}{3\times 138.5}=205.05\text{g}$ of $C{l}_2$