\(4HCl+MnO2-->Cl2+2H2O+MnCl2\)
\(n_{MnO2}=\frac{17,4}{87}=0,2\left(mol\right)\)
\(n_{Cl2}=n_{MnO2}=0,2\left(mol\right)\)
\(V_{Cl2}=0,2.22,4=4,48\left(l\right)\)
\(H\%=\frac{3,8}{4,48}.100\%=84,82\%\)
nCl2=3,8\22,4=0,17 mol
nMnO2=17,4\87=0,2 mol
=>MnO2+4HCl-->MnCl2+Cl2+H2O
....0,2-----------------------0,2 mol
=>H=0,17\0,2.100=85 %