\(n_{KMnO_4}=\dfrac{9,48}{158}=0,06\left(mol\right)\)
PT: \(2KMnO_4+16HCl_đ\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)
Theo PT: \(n_{Cl_2\left(LT\right)}=\dfrac{5}{2}n_{KMnO_4}=0,15\left(mol\right)\)
\(\Rightarrow V_{Cl_2\left(LT\right)}=0,15.22,4=3,36\left(l\right)\)
\(\Rightarrow H=\dfrac{2,52}{3,36}.100\%=75\%\)