Ta có :
\(\dfrac{m_{Pb}}{m_{Fe}}=3,696\Rightarrow\dfrac{n_{Pb}}{n_{Fe}}=3,696:\dfrac{207}{56}=\dfrac{1}{1}\)
Gọi $n_{Fe} = n_{Pb} = a(mol) \Rightarrow 207a + 56a = 52,6$
$\Rightarrow a = 0,2(mol)$
$PbO + H_2 \xrightarrow{t^o} Pb + H_2O$
$FeO + H_2 \xrightarrow{t^o} Fe + H_2O$
Theo PTHH : $n_{H_2} = n_{Pb} + n_{Fe} = 0,4(mol)$
$V_{H_2} = 0,4.22,4 = 8,96(lít)$