Câu `15:`
`n_{H_2}=\frac{4,48}{22,4}=0,2(mol)`
`a)` `Mg+2HCl\to MgCl_2+H_2`
`ZnO+2HCl\to ZnCl_2+H_2O`
`=> n_{Mg}=n_{H_2}=0,2(mol)`
`=> m_{Mg}=0,2.24=4,8g`
`=> m_{ZnO}=34-4,8=29,2g`
`=> %m_{ZnO}=\frac{29,2.100%}{34}\approx 85,9%`
`=> %m_{Mg}=\frac{4,8.100%}{34}\approx 14,1%`
`b)` `n_{ZnO}=\frac{29,2}{81}=\frac{146}{405}(mol)`
`∑n_{HCl}=2n_{ZnO}+2n_{Mg}=\frac{454}{405}(mol)`
`=> `$V_{HCl}=\dfrac{\dfrac{454}{405}}{2}\approx 0,56(l)$
