Nung nóng 48,12 gam hỗn hợp gồm Al,Al2O3 và Cr2O3 trong khí trơ,

* Hướng dẫn giải

Đáp án A

$$\begin{gathered} 24,06X\left\{\begin{array}{l}Al:n_{NaOH}-2n_{A{l}_2{O}_3}=0,34-2x\\ A{l}_2{O}_3:x\\ C{r}_2{O}_3:y\\ Cr:z\end{array}\right.\stackrel{NaOH:0,34}{\to }7,68\left\{\begin{array}{l}Cr\\ C{r}_2{O}_3\end{array}\right. \\ {m}_{Al}+m_{A{l}_2{O}_3}=24,06-7,68\to 27.\left(0,34-2x\right)+102x=16,38\to x=0,15 \\ 24,06X\left\{\begin{array}{l}Al:0,19\\ A{l}_2{O}_3:0,15\\ C{r}_2{O}_3:y\\ Cr:z\end{array}\right.\stackrel{HCl}{\to }\left\{\begin{array}{l}{H}_2:0,11\\ 61,57\left\{\begin{array}{l}A{l}^{3+}:0,34\\ C{r}^{n+}:2y+z\\ C{l}^{-}\end{array}\right.\end{array}\right. \\ 2H^{+}+O^{2-}\stackrel{}{\to }{H}_{2}O \\ {n}_{O^{2-}}=3n_{A{l}_2{O}_3}+3n_{C{r}_2{O}_3}=0,45+3y\to n_{H^{+}}=0,9+6y \\ 2H^{+}+2e\stackrel{}{\to }{H}_2 \\ \to n_{C{l}^{-}}=n_{HCl}=\sum n_{H^{+}}=0,9+6y+0,22=1,12+6y \\ \to 0,34.27+52.\left(2y+z\right)+35,5.\left(1,12+6y\right)=61,57\left(1\right) \\ {m}_{\frac{1}{2}X}=24,06\to 0,34.27+52.\left(2y+z\right)+16.\left(0,45+3y\right)=24,06\left(2\right) \\ \to \left\{\begin{array}{l}y=0,03\\ z=0,06\end{array}\right. \end{gathered}$$

Hỗn hợp ban đầu gồm: $\left\{\begin{array}{l}A{l}_2{O}_3:0,12\left(BTO\right)\\ C{r}_2{O}_3:0,06\left(BTCr\right)\\ Al:0,1\left(BTAl\right)\end{array}\right.$

Phản ứng nhiệt nhôm: $\begin{array}{l}2Al+C{r}_2{O}_3\stackrel{t°}{\to }A{l}_2{O}_3+2Cr\\ 0,10,06\\ 0,060,03\leftarrow 0,06\end{array}$

Hiệu suất phản ứng tính theo Al vì $\frac{0,1}{2}<0,06$

$H\%=\frac{0,06}{0,1}.100=60\%$