`20)`

`a)`

`n_{HCl}={175.14,6\%}/{36,5}=0,7(mol)`

Giả sử hh toàn Mg:

`->n_{hh\ max}=8/{24}=0,33(mol)`

`Mg+2HCl->MgCl_2+H_2`

`0,33<{0,7}/2->HCl` luôn dư

`->A` còn axit dư

`b)`

`X:Mg:x(mol);Fe:y(mol)`

`24x+56y=8(1)`

`Fe+2HCl->FeCl_2+H_2`

`Mg+2HCl->MgCl_2+H_2`

Theo pt:

`n_{H_2}=x+y={8-7,6}/2=0,2(2)`

`(1)(2)->x=y=0,1(mol)`

`\%m_{Mg}={0,1.24}/{80}.100\%=30\%`

`\%m_{Fe}=100-30=70\%`

`21)`

`a)`

`n_{Al}={10,8}/{27}=0,4(mol)`

`2Al+6HCl->2AlCl_3+3H_2`

`2Al+3H_2SO_4->Al_2(SO_4)_3+3H_2`

Theo pt:

`n_{H_2}=3/2n_{Al}=0,6(mol)`

`V_{H_2}=0,6.24,79=14,784(L)`

`b)`

Theo pt:

`n_{HCl}+2n_{H_2SO_4}=2n_{H_2}=1,2(mol)`

`->{m.18,25\%}/{36,5}+{2m.19,6\%}/{36,5}=1,2`

`->0,005m+0,004m=1,2`

`->m={400}/3(g)`

`c)`

`n_{HCl}={{400}/{3}.18,25\%}/{36,5}=2/3(mol)`

`->n_{AlCl_3}=1/3n_{HCl}=2/9(mol)`

`n_{H_2SO_4}={1,2-2/3}/2=4/{15}(mol)`

`->n_{Al_2(SO_4)_3}=1/3n_{H_2SO_4}=4/{45}(mol)`

`m_{dd\ sau}=10,8+{400}/{3}-0,6.2=142,93(g)`

`C\%_{AlCl_3}={2/{9}.133,5}/{142,93}.100\%=20,76\%`

`C\%_{Al_2(SO_4)_3}={4/{45}.342}/{142,93}.100\%=21,27\%`

`22)`

`a)`

`n_{Mg}=6/{24}=0,25(mol)`

`n_{HCl}=0,2.1,5=0,3(mol);n_{H_2SO_4}=0,2.0,4=0,08(mol)`

`Mg+2HCl->MgCl_2+H_2`

`Mg+H_2SO_4->MgSO_4+H_2`

`0,25>{0,3}/2+0,08->Mg` dư

`->` Dd A không làm đổi màu quỳ tím

`b)`

Theo pt:

`n_{H_2}=1/2n_{HCl}+n_{H_2SO_4}=0,23(mol)`

`V_{H_2}=0,23.24,79=5,7017(L)`

`c)`

`n_{MgCl_2}=1/2n_{HCl}=0,15(mol)`

`n_{MgSO_4}=n_{H_2SO_4}=0,08(mol)`

`C_{M\ MgCl_2}={0,15}/{0,2}=0,75M`

`C_{M\ MgSO_4}={0,08}/{0,2}=0,4M`