Cho m gam bột Fe vào 200 ml dung dịch hỗn hợp X chứa H2SO4 1M, Fe(NO3)3 0,5M và CuSO4 0,25M

* Hướng dẫn giải

$\begin{array}{l}+Sơ đồphản\text{\hspace{0.17em}\hspace{0.17em}ứ}ng:\\ \underset{m\left(g\right)}{\underbrace{Fe}}+\underset{X}{\underbrace{\left\{\begin{array}{l}{H}_{2}S{O}_4:0,2\\ Fe{\left(N{O}_3\right)}_3:0,1\\ CuS{O}_4:0,05\end{array}\right\}}}\stackrel{}{\to }\underset{\frac{0,2.2}{4}=0.1}{\underbrace{NO\uparrow }}+\underset{0,75m\left(g\right)}{\underbrace{\left\{\begin{array}{l}Fe\downarrow \\ Cu\downarrow \end{array}\right\}}}+\underset{Y}{\underbrace{\left\{\begin{array}{l}S{O_4}^{2-}:0,25\\ N{O_3}^{-}:0,3-0,1=0,2\\ F{e}^{2+}:?\end{array}\right\}}}\\ +\left\{\begin{array}{l}BTĐT:n_{F{e}^{2+}}=\frac{0,25.2+0,2}{2}=0,35\\ BTKL:m+56.0,1+0,05.64=0,75m+0,35.56\end{array}\right.\Rightarrow \boxed{m=43,2gam}\end{array}$

Chọn D