`n_{Zn}=\frac{97,5}{65}=1,5(mol)`
`a)` Phương trình:
`Zn+2HCl\to ZnCl_2+H_2`
`b)` `n_{H_2}=n_{ZnCl_2}=n_{Zn}=1,5(mol)`
`=> V_{H_2}=1,5.22,4=33,6(l)`
`m_{ZnCl_2}=1,5.136=204g`
`c)` `Fe_2O_3+3H_2\overset{t^o}{\to}2Fe+3H_2O`
`=> n_{Fe}=\frac{2}{3}n_{H_2}=1(mol)`
`=> m_{Fe}=1.56=56g`