a)
$4P + 5O_2 \xrightarrow{t^o} 2P_2O_5$
$n_P=\frac{46,5}{31}=1,5 (mol)$
$→n_{O_2}=\frac{5}{4}n_{P}=\frac{5}{4}×1,5=1,875 (mol)$
$→m_{O_2}=1,875×32=60 (g)$
b)
$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$
$n_{Al}=\frac{67,5}{27}=2,5 (mol)$
$→n_{O_2}=\frac{3}{4}n_{Al}=\frac{3}{4}×2,5=1,875 (g)$
$→m_{O_2}=1,875×32=60 (g)$
c)
$2H_2 + O_2 \xrightarrow{t^o} 2H_2O$
$n_{H_2}=\frac{33,6}{22,4}=1,5 (mol)$
$→n_{O_2}=\frac{1}{2}n_{H_2}=\frac{1}{2}×1,5=0,75 (mol)$
$→m_{O_2}=0,75×32=24 (g)$
