PTHH: MnO2 + 4HCl --> MnCl2 + Cl2 + 2H2O
0,25----->1---------------->0,25
\(V_{ddHCl}=\dfrac{1}{2}=0,5\left(l\right)=500\left(ml\right)\)
\(V_{Cl_2}=0,25.22,4=5,6\left(l\right)\)
Ta có \(n_{M_nO2}=0,25\left(mol\right)\)
\(PTPƯ:M_nO2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)
\(\Rightarrow V_{Cl2}=0,25.2,24=5,6\left(l\right)\)