\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(n_{Zn}=n_{H_2SO_4}=n_{ZnSO_4}=n_{H_2}=0,1\left(mol\right)\)
\(C\%_{H_2SO_4}=\dfrac{0,1.98}{100}.100=9,8\%\)
\(m_{ddsaupu}=6,5+100-0,1.2=106,3\left(g\right)\)
=> \(C\%_{ZnSO_4}=\dfrac{161.0,1}{106,3}.100=15,15\%\)