Hòa tan hoàn toàn 29,6 gam hỗn hợp H gồm FeS2 và CuO vào 500 gam

* Hướng dẫn giải

Đáp án A

$\begin{array}{l}+ Sơ đồ phản ứng:\\ \\ \left\{\begin{array}{l}Fe{S}_2\\ CuO\end{array}\right\} \left\{\begin{array}{l} N{O}_2\\ \left\{\begin{array}{l}F{e}^{3+}, C{u}^{2+}\\ S{O_4}^{2-}, N{O_3}^{-}\\ H^{+}\end{array}\right\}\stackrel{cô cạn}{\to }{\overbrace{\left\{\begin{array}{l}F{e}^{3+}, C{u}^{2+}\\ S{O_4}^{2-}, N{O_3}^{-}\end{array}\right\}}}^{muối chứa các ion}\end{array}\right.\\ \\ + \left\{\begin{array}{l}{n}_{Fe{S}_2}=x\\ n_{CuO}=y\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{S{O_4}^{2-}}=2x\\ n_{N{O_3}^{-}/muối}=3x+2y-4x=2y-x\end{array}\right.\\ \Rightarrow \left\{\begin{array}{l}120x+80y=29,6\\ 56x+64y+2x.96+62(2y-x)=56,12\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=0,14\\ y=0,16\end{array}\right.\\ + n_{HN{O}_3 pư}=n_{N{O}_2}+n_{N{O_3}^{-}/muối}=15n_{Fe{S}_2}+(3n_{F{e}^{3+}}+2n_{C{u}^{2+}}-2n_{S{O_4}^{2-}})=2,28\\ \Rightarrow n_{HN{O}_3 đem pư}=2,28+2,28.10\%=2,508\Rightarrow C{\%}_{HN{O}_3}\approx 31,6\%\end{array}$