Đáp án:
`m_{dd HCl}=50gam`
`m_{H_2}=0,1gam`
`V_{H_2}=1,12 lít`
Giải thích các bước giải:
`m_{Cu}=3,2(gam)`
`-> m_{Mg}=4,4-3,2=1,2(gam)`
`-> n_{Mg}=(1,2)/(24)=0,05(mol)`
`Mg + 2HCl -> MgCl_2 + H_2`
Theo PTHH:
`n_{HCl}=2.n_{Mg}=2.0,05=0,1(mol)`
`-> m_{HCl}=0,1.36,5=3,65(gam)`
`=> m_{dd HCl}=3,65 : 7,3%=50(gam)`
Theo PTHH:
`n_{H_2}=n_{Mg}=0,05(mol)`
`-> m_{H_2}=0,05.2=0,1(gam)`
`-> V_{H_2}=0,05.22,4=1,12(lít)`