\(n_{Zn}=\frac{13}{65}=0,2\left(mol\right)\)
a. \(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\)
b. \(n_{HCl}=n_{Zn}=0,2\left(mol\right)\)
\(\rightarrow m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(\rightarrow m_{dd_{HCl}}=\frac{7,3}{20\%=36,5\left(g\right)}\)
\(n_{H2}=n_{Zn}=0,2\left(mol\right)\)
\(\rightarrow\) Khối lượng dd sau phản ứng:
m=mZn+mddHCl-mH2\(=13-36,5-0,2.2=49,1\)
\(C\%_{ZnCl2}=\frac{0,2.136}{49,1}=55,4\%\)