1.
$\Delta m=7g=m_{KL}-m_{H_2}$
$\Rightarrow m_{H_2}=7,8-7=0,8g$
$\Rightarrow n_{H_2}=\dfrac{0,8}{2}=0,4(mol)$
Bảo toàn nguyên tố:
$n_{Cl^-}=n_{HCl}=2n_{H_2}=0,8(mol)$
$\to m_{\text{muối}}=m_{KL}+m_{Cl^-}=7,8+0,8.35,5=36,2g$
2.
Gọi $x$, $y$ là số mol $MnO_2$, $KMnO_4$
$\Rightarrow 87x+158y=41,9$ $(1)$
$n_{Mn}=\dfrac{41,9.52,506\%}{55}=0,4(mol)$
$\Rightarrow x+y=0,4$ $(2)$
$(1)(2)\Rightarrow x=0,3; y=0,1$
Bảo toàn e:
$2n_{MnO_2}+5n_{KMnO_4}=2n_{Cl_2}$
$\Rightarrow n_{Cl_2}=\dfrac{2x+5y}{2}=0,55(mol)$
$\to V=0,55.22,4=12,32l$