From the following reaction sequence

$C{l}_2 +2KOH\to KClO+KCl$

$$\begin{gathered} 3KClO\to 2KCl+KCl{O}_3 \\ 4KCl{O}_3\to 3KCl{O}_4+KCl \end{gathered}$$

Calculate the mass of chlorine needed to produce 100g of $KCl{O}_4$.

Rewriting the given equations,

$$\begin{gathered} 12C{l}_2+24KOH\to 12KCl+12KClO+12H_{2}O \\ 12KClO\to 8KCl+4KCl{O}_3 \\ 4KCl{O}_3\to KCl+3KCl{O}_4 \end{gathered}$$

Adding all the equations

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

Molecular weight =138.5

Applying  mole-mole analysis,

3 moles of $KCl{O}_4$is obtained from 12 moles of $C{l}_2$

$\therefore$$\frac{100}{138.5}$moles of $KCl{O}_3$will be obtained from $\frac{12\times 100}{3\times 138.5}=2.88$moles of $C{l}_2$$=2.888\times 71 = 205.05$gm of  $C{l}_2$