Giúp em với ạ.........

Tham khảo :

$n_{FeS_2}=a(mol);n_{Cu_2S}=b(mol)$

$4FeS_2 +11O_2 \xrightarrow{t^o}2Fe_2O_3+8SO_2$

$2a$               `←`              $a(mol)$

$Cu_2S+2O_2 \xrightarrow{t^o}2CuO+SO_2$

$\dfrac{b}{2}$           `←`            $b(mol)$

$m_{Fe_2O_3}=60\%m_{oxide}$

$m_{CuO}=40\%m_{oxide}$

`->`$\dfrac{m_{Fe_2O_3}}{m_{CuO}}=\dfrac{60\%m_{oxide}}{40\%m_{oxide}}=\dfrac{3}{2}$

`->`$\dfrac{160a}{80b}=\dfrac{3}{2}$

`->`$320a=240b$

`=>`$b=\dfrac{4}{3}a$

`=>`$\begin{cases} n_{Cu_2S}=\dfrac{1}{2} \cdot \dfrac{4}{3}a=\dfrac{2}{3}a\\n_{FeS_2}=2a \end{cases}$

`=>`$\%n_{Cu_2S}=\dfrac{\dfrac{2}{3}a}{\dfrac{2}{3}a+2a}.100\%=25\%$

Cho `m_{o x i t}=100(g)`

`->m_{Fe_2O_3}=60(g);m_{CuO}=40(g)`

`->n_{Fe_2O_3}={60}/{160}=0,375(mol);n_{CuO}={40}/{80}=0,5(mol)`

`4FeS_2+11O_2`  $\xrightarrow{t^o}$  `2Fe_2O_3+8SO_2`

`0,75←2,0625->0,375(mol)`

`Cu_2S+2O_2`  $\xrightarrow{t^o}$  `2CuO+SO_2`

`0,25←0,5←0,5(mol)`

`->\%n_{Cu_2S}={0,25}/{0,25+0,75}.100\%=25\%`