Đáp án:
Giải thích các bước giải:
`Mg+2HCl→MgCl_2+H_2↑`
`n_{Mg}=\frac{4,8}{24}=0,2\ mol`
a) `n_{H_2}=n_{Mg}=0,2\ mol`
`V_{H_2}=0,2.22,4=4,48\ lít`
b) `n_{HCl}=2n_{Mg}=0,4\ mol`
`m_{HCl}=0,4.36,5=14,6\ gam`
`m_{dd\ HCl}=\frac{m_{HCl}}{C%}=\frac{14,6}{10%}=146\ gam`
`m_{dd}=m_{Mg}+m_{dd\ HCl}-m_{H_2}`
`m_{dd}=4,8+146-(0,2.2)=150,4\ gam`
`C%_{MgCl_2}=\frac{0,2.(24+71)}{150,4} \approx 12,63%`