_ \(n_{Zn}=\dfrac{13}{65}=0,2mol\)
Theo PTHH : \(n_{ZnCl_2}=n_{Zn}=n_{H_2}=0,2mol\)
\(\Rightarrow m_{ZnCl_2}=0,2.136\left(g\right)\)
\(V_{H_2}=0,2.22,4=4,48\left(l\right).\)
Zn +2HCl->ZnCl2+H2
nZn=13/65=0,2 mol
theo phương trình phản ứng:
nZnCl\(_2\)=nH\(_2\)=nZn=0,2 mol
mZnCl\(_{2_{ }}\)=0,2. 136=27,2gam
VH\(_2\)=0,2.22,4=4,48 lít