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

`n_{H_2} = (1,2395)/(24,79) = 0,05` mol

a)Theo PT

`n_{Al_2(SO_4)_3} = 1/3n_{H_2} = 1/3 . 0,05 = (1)/(60)` mol

`=>m_{Al_2(SO_4)_3} = 1/(60) . 342 = 5,7g`

b)Theo PT

`n_{Al} = 2/3n_{H_2} = 2/3 . 0,05 = 1/(30)` mol

`=>%m_{Al} = ((1)/(30). 27)/(12) . 100% = 7,5%`

`%m_{Ag} = 100% - 7,5% = 92,5%`

c)

`Ag` không tác dụng với `O_2`

`4Al + 3O_2 \overset(t^°)(->) 2Al_2O_3`

`n_{O_2} = (9,6)/(32) = 0,3` mol

Theo PT

`n_{Al} = 4/3n_{O_2} = 4/3 . 0,3 = 0,4` mol

`=>%m_{Al} = (0,4 . 27)/(12) . 100% = 90%`

`%m_{Ag} = 100% - 90% = 10%`

$a)\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ n_{H_2}=\dfrac{1,2395}{24,79}=0,05(mol)\\ n_{Al_2(SO_4)_3}=\dfrac{1}{3}n_{H_2}=\dfrac{1}{60}(mol)\\ m_{Al_2(SO_4)_3}=\dfrac{1}{60}.342=5,7(g)\\ b)\\ n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{1}{30}(mol)\\ \%m_{Al}=\dfrac{\frac{1}{30}.27}{12}.100\%=7,5\%\\ \%m_{Ag}=100\%-7,5\%=92,5\%\\ c)\\ 4Al+3O_2\xrightarrow{t^o} 2Al_2O_3\\  n_{O_2}=\dfrac{9,6}{32}=0,3(mol)\\ n_{Al}=\dfrac{4}{3}n_{O_2}=0,4(mol)\\ \%m_{Al}=\dfrac{0,4.27}{12}.100\%=90\%\\ \%m_{Ag}=100\%-90\%=10\%$