Cho 5,4 gam Al vào dung dịch chứa 0,2 mol

Đáp án B

$n_{Al}=0,2mol$

Phương trình ion:

 $\begin{array}{l}Al+3F{e}^{3+}\to A{l}^{3+}+3F{e}^{3+}\\ \frac{0,2}{3}\leftarrow 0,2\to \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}0,2mol\\ Al+3H^{+}\to A{l}^{3+}+\frac{3}{2}{H}_2\\ 0,1\leftarrow 0,3\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}mol\\ 2Al+3F{e}^{2+}\to 2A{l}^{3+}+3Fe\\ \frac{1}{30}0,2\to \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}0,05mol\end{array}$

Do đó: $m_{cr}=m_{Fe}=0,05.56=2,8gam$ .