Gọi 3x, 2x là số mol Fe, Zn.
$\Rightarrow 56.3x+65.2x=2,98$
$\Leftrightarrow x=0,01$
$\to n_{Fe}=0,02(mol); n_{Zn}=0,03(mol)$
Bảo toàn e: $n_{H_2}=\dfrac{2n_{Fe}+2n_{Zn}}{2}=0,05(mol)$
$\Rightarrow V_{H_2}=0,05.22,4=1,12l$
$n_{HCl}=2n_{H_2}=0,1(mol)$
$\Rightarrow m_{HCl}=0,1.36,5=3,65g$
Bảo toàn khối lượng:
$m_{\text{muối}}=2,98+3,65-0,05.2=6,53g$