\(n_{Cl_2}=\dfrac{7.1}{71}=0.1\left(mol\right)\)
\(MnO_2+4HCl_{\left(đ\right)}\rightarrow MnCl_2+Cl_2+2H_2O\)
\(0.1................................................0.1\)
\(n_{MnO_2\left(tt\right)}=\dfrac{0.1}{80\%}=0.125\left(mol\right)\)
\(m_{MnO_2}=0.125\cdot87=10.875\left(g\right)\)