Đáp án:
a)
`Zn + 2HCl -> ZnCl_2 + H_2`
b)
`n_{Zn} = (9,7)/(65) = (97)/(650)` `mol`
`n_{ZnCl_2} = n_{Zn} = (97)/(650)` `mol`
`m_{ZnCl_2} = (97)/(650) . 136 = 20,3` `gam`
c)
`n_{H_2} = n_{Zn} = (97)/(650)` `mol`
`V_{H_2} = (97)/(650) . 22,4 = 3,343` `l`
d)
`n_{HCl} = 2 . n_{Zn} = 2 . (97)/(650) = (97)/(325)` `mol`
`C_{M_(HCl)} = [(97)/(325)]/(0,1) = 2,985` `M`