$n_{Mg}=\dfrac{3,6}{24}=0,15(mol)$
PTHH: $2Mg+O_2$ $\frac{t^{\circ}}{}$> $2MgO$
Theo PTHH:
$n_{O_2}=\dfrac{1}{2}n_{Mg}=\dfrac{1}{2}.0,15=0,075(mol)$
$V_{O_2}=0,075.22,4=1,68(l)$
$V_{kk}=1,68:20\%=8,4(l)$
$→\dfrac{V_{kk}}{V_{O_2}}=\dfrac{8,4}{1,68}=5$