\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có :
\(b,n_{Zn}=\frac{13}{65}=0,2\left(mol\right)\)
Theo phương trình : \(n_{H2}=n_{Zn}=0,2\left(mol\right)\)
\(\rightarrow V_{H2}=0,2.22,4=4,48\left(l\right)\)
\(c,n_{HCl}=2.n_{HCl}=0,2.2=0,4\left(mol\right)\)
\(\rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)