Mn ơi giải giúp mình với ạaa plss Mình cảm ơn nhiều ạaaa

1.

`a)`

`n_{H_2}={6,72}/{22,4}=0,3(mol)`

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

`Al_2O_3+6HCl->2AlCl_3+3H_2O`

Theo PT: `n_{Al}=2/3n_{H_2}=0,2(mol)`

`->\%m_{Al}={0,2.27}/{36}.100\%=15\%`

`->\%m_{Al_2O_3}=100-15=85\%`

`b)`

`n_{Al_2O_3}={36-0,2.27}/{102}=0,3(mol)`

Theo PT: `n_{HCl}=6n_{Al_2O_3}+3n_{Al}=2,4(mol)`

`200ml=0,2l`

`->C_{M\ HCl}={2,4}/{0,2}=12M`

`c)`

Theo PT: `n_{AlCl_3}=1/3n_{HCl}=0,8(mol)`

`->C_{M\ AlCl_3}={0,8}/{0,2}=4M`

2.

`a)`

`Zn+H_2SO_4->ZnSO_4+H_2`

Theo PT: `n_{Zn}=n_{H_2}={4,48}/{22,4}=0,2(mol)`

`->\%m_{Zn}={0,2.65}/{32,2}.100\%\approx 40,37\%`

`->\%m_{Cu}=100-40,37=59,63\%`

`b)`

Theo PT: `n_{H_2SO_4}=n_{Zn}=0,2(mol)`

`->C\%_{H_2SO_4}={0,2.98}/{200}.100\%=9,8\%`

`c)`

Theo PT: `n_{ZnSO_4}=n_{Zn}=0,2(mol)`

`->C\%_{ZnSO_4}={0,2.161}/{200+0,2.65-0,2.2}.100\%\approx 15,146\%`

$\rm 1)\\a)\\n_{H_2}=6,72:22,4=0,3(mol)\\2Al+6HCl\xrightarrow[]{}2AlCl_3+3H_2(1)\\Al_2O_3+6HCl\xrightarrow[]{}2AlCl_3+3H_2O(2)\\\xrightarrow[]{(1)}n_{Al}=(2/3)n_{H_2}=0,2(mol)\\\xrightarrow[]{}m_{Al}=0,2.27=5,4(g)\\\xrightarrow[]{}\\\%m_{Al}=(5,4:36).100\%=15\%\\\%m_{Al_2O_3}=100\%-15\%=85\%\\b)\\m_{Al_2O_3}=36.85\%=30,6(g)\\\xrightarrow[]{}n_{Al_2O_3}=30,6:102=0,3(mol)\\\xrightarrow[]{(1)}n_{HCl}=3n_{Al}=0,6(mol)\\\xrightarrow[]{(2)}n_{HCl}=6n_{Al_2O_3}=1,8(mol)\\\xrightarrow[]{}Tổng-n_{HCl}=0,6+1,8=2,4(mol)\\Đổi:200(ml)=0,2(l)\\\xrightarrow[]{}C_{M_{dd_{HCl}}}=2,4:0,2=12M\\c)\\\xrightarrow[]{(1)(2)}n_{AlCl_3}=(1/3)n_{HCl}=0,8(mol)\\\xrightarrow[]{}C_{M_{dd_{AlCl_3}}}=0,8:0,2=4M\\2)\\a)\\n_{H_2}=4,48:22,4=0,2(mol)\\Zn+H_2SO_4\xrightarrow[]{}ZnSO_4+H_2\\\xrightarrow[]{PTHH}n_{Zn}=n_{H_2}=0,2(mol)\\\xrightarrow[]{}m_{Zn}=0,2.65=13(g)\\\xrightarrow[]{}\\\%m_{Zn}=(13:32,2).100\%=40,37\%\\\%m_{Cu}=100\%-40,37\%=59,63\%\\b)\\\xrightarrow[]{PTHH}n_{H_2SO_4}=n_{H_2}=0,2(mol)\\\xrightarrow[]{}m_{H_2SO_4}=0,2.98=19,6(g)\\\xrightarrow[]{}C\%_{dd_{H_2SO_4}}=(19,6:200).100\%=9,8\%\\c)\\\xrightarrow[]{PTHH}n_{ZnSO_4}=n_{H_2}=0,2(mol)\\\xrightarrow[]{}C\%_{dd_{ZnSO_4}}=[(0,2.161):(13+200-0,2.2)].100\%=15,15\%$