Hòa tan hết 2,16 gam hỗn hợp Cr và Fe trong dung dịch HCl (loãng),

Đáp án B.

$n_{{\text{H}}_2}=0,04 \text{mol}$

Ta có

$$\begin{gathered} \begin{array}{l}\text{Fe}\to \stackrel{+2}{\text{Fe}}+2\text{e}\\ \text{x\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2x}\end{array} \begin{array}{l}2\stackrel{+1}{\text{H}}+2\text{e}\to {\stackrel{\text{0}}{\text{H}}}_{\text{2}}\\ 0,08 0,04\end{array} \\ \begin{array}{l}\text{Cr}\to \stackrel{+2}{\text{Cr}}+2\text{e}\\ y2y\end{array} \\ \text{BTe: }{n}_{e(+)}=n_{e(-)}\Rightarrow 2x+2y=0,08 \\ 52x+56y=2,16 \\ \Rightarrow x=y=0,02 \\ {m}_{\text{Cr}}=0,02.52=1,04( \text{g}) \end{gathered}$$