PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
1) Ta có: \(n_{Zn}=\frac{13}{65}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,4mol\\n_{ZnCl_2}=0,2mol\\n_{H_2}=0,2mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{HCl}=0,2\left(l\right)=200\left(ml\right)\\m_{ZnCl_2}=27,2\left(g\right)\\V_{H_2}=4,48\left(l\right)\end{matrix}\right.\)
2) Ta có: \(C_{M_{ZnCl_2}}=\frac{0,2}{0,2}=1\left(M\right)\) (Coi như thể tích dd không đổi)
3) Ta có: \(\left\{{}\begin{matrix}m_{ddHCl}=200\cdot1,1=220\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{dd}=m_{Zn}+m_{ddHCl}-m_{H_2}=13+220-0,4=232,6\left(g\right)\)
\(\Rightarrow C\%_{ZnCl_2}=\frac{27,2}{232,6}\cdot100\approx11,72\%\)