Nung nóng 19,52 gam hỗn hợp gồm...

B

- Hướng dẫn giải

Đáp án B

+ Sơ đồ phản ứng:

$$\begin{gathered} \underset{19,52 gam}{\underbrace{\left\{ Al \\ C{r}_2{O}_3\right\}}}\stackrel{t^o}{\to }\underset{hỗn hợp X}{\underbrace{\left\{Al, A{l}_2{O}_3 \\ Cr, C{r}_2{O}_3\right\}}} \underset{0,96 mol}{\overset{HCl}{\to }}\left\{\begin{array}{l}{H}_2\uparrow : 0,18 mol\\ \underset{dung dịch Y}{\underbrace{\left\{AlC{l}_3, CrC{l}_3 \\ CrC{l}_2\right\}}}\stackrel{NaOH dư}{\to }Cr(OH{)}_2\downarrow \end{array}\right. \\ + \left\{BTNT H: \underset{0,96}{\underbrace{n_{HCl}}}=2\underset{?}{\underbrace{n_{H_{2}O}}}+2\underset{0,18}{\underbrace{n_{H_2}}} \\ BTNT O: 3\underset{?}{\underbrace{n_{C{r}_2{O}_3 ban đầu}}}=\underset{?}{\underbrace{n_{H_{2}O}}}\right.\Rightarrow \left\{n_{H_{2}O}=0,3; n_{C{r}_2{O}_3 ban đầu}=0,1 \\ {n}_{Al bđ}=\frac{19,52-0,1.152}{27}=0,16\right. \\ + \left\{n_{C{r}_2{O}_3 pư}=x \\ {n}_{Al p¨o}=2x\right.\Rightarrow \left\{n_{Cr tạo thành}=2x \\ {n}_{Al dư}=0,16-2x\right.\Rightarrow \left\{BTE: 2x.2+3(0,16-2x)=2.0,18 \\ BTNT Cr: n_{Cr{\left(OH\right)}_2}=n_{Cr}=2x\right. \\ \Rightarrow \left\{x=0,06 \\ {n}_{Cr{\left(OH\right)}_2}=0,12\right.\Rightarrow \boxed{m_{Cr{\left(OH\right)}_2}=10,32 gam} \end{gathered}$$