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Trả lời:
$Pt:\,MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O$
$n_{MnO_2}=\dfrac{34,8}{87}=0,4\,(mol)$
$⇒n_{Cl_2}=n_{MnO_2}=0,4\,(mol)$
$⇒V_{Cl_2}=0,4.22,4=8,96\,(l)$
$n_{MnCl_2}=n_{MnO_2}=0,4\,(mol)$
$⇒m_{MnCl_2}=0,4.126=50,4\,(g)$.