`Zn+2HCL -> ZnCl_2 + H_2 `
a/ `n_(Zn)=32,5/65=0,5` (mol)
`n_(HCl)=47,45/36,5=1,3` (mol)
Lập tỉ lệ :
`n_(Zn) <(n_(HCl))/2`
`=>Zn` phản ứng hết
Theo PTHH : `n_(H_2)=n_(Zn)=0,5` (mol)
`=>V_(H_2)=0,5.22,4=11,2` (l)
b/
Theo PTHH
`n_(ZnCl_2)=n_(Zn)=0,5` (mol)
`=>m_(ZnCl_2)=0,5.136=68` (g)