a) $m_{CuO} = 16 \times 25$%$=4 (g) ⇒ n_{CuO} = \frac{4}{80} = 0.05(mol)$
$m_{Fe_{2}O_{3}} = 16 - 4 = 12 (g) ⇒n_{Fe_{2}O_{3}} = \frac{12}{160} = 0.075(mol)$
PTHH: $Fe_{2}O_{3} + 3H_{2} \xrightarrow{t^o} 2Fe + 3H_{2}O$ (1)
$CuO + H_{2} \xrightarrow{t^o} Cu + H_{2}O$ (2)
Theo PTHH (1): $n_{Fe} = 2n_{Fe_{2}O_{3}} = 0.15 (mol)$
⇒ $m_{Fe} = 0.15 \times 56 = 8.4 (g)$
Theo PTHH (2): $n_{Cu} = n_{CuO} = 0.05 (mol)$
⇒ $m_{Cu} = 0.05 \times 64 = 3.2 (g)$
b) Theo PTHH (1): $n_{H_{2}(1)} = 3n_{Fe_{2}O_{3}} = 0.225 (mol)$
Theo PTHH (2): $n_{H_{2}(2)} = n_{CuO} = 0.05 (mol)$
⇒ $n_{H_{2}(tổng)} = 0.225 + 0.05 = 0.275 (mol)$
⇒ $V_{H_{2}(tổng)} = 0.275 \times 22.4 = 6.16 (l)$