\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(m_{H_2SO_4}=\dfrac{10.196}{100}=19,6\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
PTHH:_______\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
Theo PT:mol:___1..........1..................1.............1
Theo ĐB:mol:___0,1.......0,2...............................
\(\Rightarrow H_2SO_4\)dư,Zn pứ hết
Theo PT: \(n_{H_2SO_4pư}=n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4dư}=0,2-0,1=0,1\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,1.98=9,8;m_{ZnSO_4}=0,1.161=16,1\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4}=\dfrac{9,8}{196+6,5}.100\%\approx4,84\%\)
\(\Rightarrow C\%_{ZnSO_4}=\dfrac{16,1}{196+6,5}.100\%\approx7,95\%\)