`n_{H_2}={5,6}/{22,4}=0,25(mol)`

`PTHH:Fe+2HCl->FeCl_2+H_2`

`Cu` không pư`

`a.n_{Fe}=n_{H_2}=0,25(mol)`

`m_{Fe}=0,25.56=14(g)`

`%Fe=14/58 .100%~~24,14%`

`%Cu=100-24,14=75,86%`

`b.n_{HCl}=2n_{H_2}=0,25.2=0,5(mol)`

`m=m_{dd HCl}={0,5.36,5.100}/10=182,5(g)`

`c.PTHH:HCl+NaOH->NaCl+H_2O`

`(mol)`    `0,5->`   `0,5`

`m_{dd NaOH}={0,5.40.100}/20=100(g)`

`a)`

`n_{H_{2}}=\frac{5,6}{22,4}=0,25(mol)`

`Fe+2HCl→FeCl_{2}+H_{2}(1)`

`0,25`                                  `0,25`          / `mol`

`⇒m_{Fe}=0,25.56=14(g)`

`⇒m_{Cu}=m_{hh}-m_{Fe}=58-14=44(g)`

`b)` Theo phương trình `(1)⇒n_{HCl}=2n_{H_{2}}=2.0,25=0,5(mol)`

`⇒m_{HCl}=0,5.36,5=18,25(g)`

`⇒m_{ddHCl}=18,25:10%=182,5(g)`

`c)`

`HCl+NaOH→NaCl+H_{2}O`

`0,5`          `0,5`                            / `mol`

`⇒m_{NaOH}=0,5.40=20(g)`

`⇒m_{ddNaOH}=20:20%=100(g)`