a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, Ta có: \(n_{H_2}=0,15\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,15.65=9,75\left(g\right)\)
c, Ta có: m dd sau pư = 9,75 + 200 - 0,15.2 = 209,45 (g)
Theo PT: \(n_{ZnCl_2}=n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow C\%_{ZnCl_2}=\dfrac{0,15.136}{209,45}.100\%\approx9,74\%\)