\(n_{H_2}=\dfrac{8,6765}{24,79}=0,35\left(mol\right)\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
0,35 0,35 0,35 0,35 ( mol )
\(m_{dd_{H_2SO_4}}=\dfrac{0,35.98}{25\%}=137,2\left(g\right)\)
\(m_{ddspứ}=0,35.65+137,2-0,35.2=159,25\left(g\right)\)
\(C\%_{ZnSO_4}=\dfrac{0,35.161}{159,25}.100=35,38\%\)