a) $n_{Mg} = \dfrac{12}{24} = 0,5(mol)$
$2Mg + O_2 \xrightarrow{t^o} 2MgO$
Theo PTHH :
$n_{O_2} = \dfrac{1}{2}n_{Mg} = 0,25(mol)$
$m_{O_2} = 0,25.32 = 8(gam)$
$V_{O_2} = 0,25.22,4 = 5,6(lít)$
b)
$n_{MgO} = n_{Mg} = 0,5(mol)$
$m_{MgO} = 0,5.40 = 20(gam)$