Bạn tham khảo:
$\underbrace{Fe}_{5,6(g)}+\underbrace{O_2}_{V=?(l)}\xrightarrow{t^{o}}\underbrace{Fe_3O_4}_{m=?(g)}$
$3Fe+2O_2 \xrightarrow{t^{o}} Fe_3O_4$
$n_{Fe}=0,1(mol)$
$a/$
$n_{Fe_3O_4}=\frac{0,1}{3}=\frac{1}{30}(mol)$
$m_{Fe_3O_4}=\frac{1}{30}.232=\frac{116}{15}(g)$
$b/$
$n_{O_2}=\frac{2}{3}.0,1=\frac{1}{15}(mol)$
$V_{O_2}=\frac{1}{15}.22,4=\frac{112}{75}(l)$