Câu `8:`
`n_{H_2}=\frac{8,96}{22,4}=0,4(mol)`
`CuO+H_2SO_4\to CuSO_4+H_2O`
`Mg+H_2SO_4\to MgSO_4+H_2`
`n_{Mg}=n_{H_2}=0,4(mol)`
`=> m_{Mg}=0,4.24=9,6g`
`%m_{Mg}=\frac{9,6.100%}{24,2}\approx 39,67%`
`%m_{CuO}=100%-39,67%=60,33%`
Câu `9:`
`n_{H_2}=\frac{2,688}{22,4}=0,12(mol)`
Cho $2,52(g) \begin{cases} Mg: x(mol)\\ Al: y(mol)\\\end{cases}$
`=> 24x+27y=2,52g(1)`
`Mg+H_2SO_4\to MgSO_4+H_2`
`2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2`
`=> x+1,5y=0,12(mol)(2)`
`(1),(2)=> x=0,06(mol), y=0,04(mol)`
`a)` `m_{Mg}=0,06.24=1,44g`
`m_{Al}=0,04.27=1,08g`
`b)` Cho chất khử duy nhất được tạo thành là `R`
BTe:
$\mathop{Mg}\limits^{0}\to \mathop{Mg}\limits^{+2}+2e$
$\mathop{Al}\limits^{0}\to \mathop{Al}\limits^{+3}+3e$
$\mathop{S}\limits^{+6}+(6-n)e\to \mathop{R}\limits^{+n} $
`=> 2n_{Mg}+3n_{Al}=(6-n).0,03`
`=> 0,06.2+0,04.3=0,03(6-n)`
`=> 0,24-0,18=-0,03n`
`=>0,06=-0,03n`
`=> n=-2`
`=>` Bạn xem lại đề
