`n_Y=(3, 136) /(22, 4) =0, 14(mol) ; M_Y=12, 25.4=49(g//mol) `

Gọi `n_{SO_2}=a;n_{H_2S}=b`

$\Rightarrow \begin{cases} a+b=0,14\\64a+34b=0, 14.49\end{cases} \Rightarrow a=b=0,07$

Ta có:`27n_{Al}+64n_{Cu}=18, 26`

Bảo toàn e: `3n_{Al}+2n_{Cu}=0, 07.2+0, 07.8`

Giải hệ `=> n_{Al}=0, 06(mol);n_{Cu}=0, 26(mol) `

`=> %m_{Al}=(0, 06.27) /(18, 26).100%=8, 87% ;%m_{Cu}=100-8, 87=91, 13%`

`M_Y=12,25.4=49(g//mol)`

`->{n_{SO_2}}/{n_{H_2S}}={49-34}/{64-34}=1`

Mà `n_{SO_2}+n_{H_2S}={3,136}/{22,4}=0,14(mol)`

`->n_{SO_2}=n_{H_2S}=0,07(mol)`

Đặt `n_{Al}=x(mol);n_{Cu}=y(mol)`

`->27x+64y=18,26(1)`

Bảo toàn e: `3x+2y=2n_{SO_2}+8n_{H_2S}=0,7(2)`

`(1)(2)->x=0,06;y=0,26`

`\%m_{Al}={0,06.27}/{18,26}.100\%\approx 8,87\%`

`\%m_{Cu}=100-8,87=91,13\%`