Ta có: $n_{H_{2}SO_{4}}=\frac{200.0,2}{1000}=0,04(mol)$
$n_{Fe}=\frac{0,56}{56}=0,01(mol)$
PTHH: $Fe + H_{2}SO_{4} → FeSO{4}+H_{2}$
(mol) 0,01 : 0,04 : 0,01 : 0,01
Vì $\frac{0,01}{1} < \frac{0,04}{1}$ nên Fe hết, $H_{2}SO_{4}$ dư
Theo PTHH:
$\left \{ {{n_{muối}=0,01(mol)} \atop {n_{H_{2}}}=0,01(mol)} \right.$ <=> $\left \{ {V_{H_{2}}=0,01.22,4=0,224(l) \atop {m_{muối}}=0,01.152=1,52(g)} \right.$