$a\big)Zn+H_2SO_4\to ZnSO_4+H_2$
$b\big)$
Theo PT: $n_{H_2}=n_{Zn}=\frac{13}{65}=0,2(mol)$
$\to V_{H_2(đktc)}=0,2\times 22,4=4,48(l)$
$c\big)$
$Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O$
Theo PT: $n_{Fe}=\frac{2}{3}n_{H_2}=\frac{2}{15}(mol)$
$\to m_{Fe}=\frac{2}{15}\times 56\approx 7,47(g)$