Hướng dẫn giải:

n_CuSO4 = 0,5.0,4 = 0,2 mol

2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu

2x       3x                    x          3x (Mol)

Theo bài ta có:

m_{Cu bám vào} – m_{Al tan} = m_{Al tăng}

⇔ 3x.64 - 2x.27 = 51,38 – 50

⇔ 138x = 1,38

⇔ x = 0,01 mol

$\Rightarrow \left\{ \begin{align}  & {{n}_{CuS{{O}_{4}}pu}}=3.x=0,03\text{ mol} \\ & {{n}_{A{{l}_{2}}{{(S{{O}_{4}})}_{3}}}}=x=0,01mol \\\end{align} \right.$$\Rightarrow \left\{ \begin{align}  & {{n}_{CuS{{O}_{4}}du}}={{n}_{CuS{{O}_{4}}}}-{{n}_{CuS{{O}_{4}}pu}}=0,17\text{ mol} \\ & {{n}_{A{{l}_{2}}{{(S{{O}_{4}})}_{3}}}}=x=0,01mol \\\end{align} \right.$

$\Rightarrow \left\{ \begin{align}  & {{C}_{M}}_{CuS{{O}_{4}}du}=\frac{0,17}{0,2}=0,85M \\ & {{C}_{M}}_{A{{l}_{2}}{{(S{{O}_{4}})}_{3}}}=x=0,05M \\\end{align} \right.$

⇒ Chọn D.