If 6 moles of $KCl{O}_3$  are decomposed according to following reactions calculate the moles of $KCl{O}_4$  produced if mole of $O_2$  produced are 3?

$$\begin{gathered} KCl{O}_3\to KCl+\frac{3}{2}{O}_2\mathrm{.....}\left(i\right) \\ 4KCl{O}_3\to 3KCl{O}_4+KCl\mathrm{.....}\left(ii\right) \end{gathered}$$
 
 
 

Let x-mole $KCl{O}_3$  reacts in reaction (i) and y mole $KCl{O}_3$ reacts in reaction (ii)
$KCl{O}_3\to KCl+\frac{3}{2}{O}_2\mathrm{....}\left(i\right)$ 
 $xmole\frac{3x}{2}mole$
 $4KCl{O}_3\to 3KCl{O}_4+KCl$
y mole
From question $x+y=6$         
and $\frac{3x}{2}=3$
 $\therefore x=2moleandy=4mole$
It means4 mole $KCl{O}_3$ reacts in reaction (ii)
From reaction (ii) $4moleKCl{O}_3$ gives $3moleKCl{O}_4$