Hoà tan hoàn toàn m gam hỗn hợp X gồm Mg, Fe, FeCO3, Cu(NO3)2 vào dung dịch chứa H2SO4 và 0,045 mol NaNO3, thu được dung dịch Y chỉ chứa 62,605 gam muối trung hoà

* Hướng dẫn giải

$\underset{m\left(g\right)X}{\underbrace{Mg,Fe,FeC{O}_3,Cu{\left(N{O}_3\right)}_2}}+\underset{dungdòchhoãnhôïp}{\underbrace{H_{2}S{O}_4,\stackrel{0,045mol}{\overbrace{NaN{O}_3}}}}\to \underset{62,605\left(g\right)Y}{\underbrace{M{g}^{2+},F{e}^{a+},C{u}^{2+},\stackrel{0,045mol}{\overbrace{N{a}^{+}}},N{H}_4^{+},S{O}_4^{2-}}}+\underset{0,17molhoãnhôïpZ}{\underbrace{\stackrel{0,02mol}{\overbrace{H_2}},C{O}_2,NO}}$

-          Cho $\underset{62,605\left(g\right)Y}{\underbrace{M{g}^{2+},F{e}^{a+},C{u}^{2+},N{H}_4^{+},S{O}_4^{2-}}}\stackrel{NaOH}{\to }\underset{31,72\left(g\right)\downarrow }{\underbrace{Fe{\left(OH\right)}_a,Cu{\left(OH\right)}_2,Mg{\left(OH\right)}_2}}+N{a}_{2}S{O}_4$

$$\begin{gathered} \Rightarrow a{n}_{F{e}^{a+}}+2n_{M{g}^{2+}}+2n_{C{u}^{2+}}+n_{N{H}_4^{+}}=n_{NaOH}=0,865\left(1\right) \\ \stackrel{BTDT\left(Y\right)}{\to }{n}_{H_{2}S{O}_4}=n_{S{O}_4^{2-}}=0,455mol \\ \Rightarrow m_{\downarrow \max }=56n_{F{e}^{a+}}+24n_{M{g}^{2+}}+64n_{C{u}^{2+}}+17\left(n_{O{H}^{-}}-n_{N{H}_4^{+}}\right)\to 56n_{F{e}^{a+}}+24n_{M{g}^{2+}}+64n_{C{u}^{2+}}=17,015+17n_{N{H}_4^{+}} \end{gathered}$$

-          Ta có: $m_Y=56n_{F{e}^{a+}}+24n_{M{g}^{2+}}+64n_{C{u}^{2+}}+23n_{N{a}^{+}}+18n_{N{H}_4^{+}}+96n_{S{O}_4^{2-}}$$\to 62,605=17,015+17n_{N{H}_4^{+}}+23.0,045+18n_{N{H}_4^{+}}+96.0,455\to n_{N{H}_4^{+}}=0,025\left(mol\right)$

$$\begin{gathered} \stackrel{BT:H}{\to }{n}_{H_{2}O}=\frac{2n_{H_{2}S{O}_4}-4n_{N{H}_4^{+}}-2n_{H_2}}{2}=0,385mol \\ \stackrel{BTKL}{\to }{m}_X=m_Y+m_Z+18n_{H_{2}O}-85n_{NaN{O}_3}-98n_{H_{2}S{O}_4}=27,2\left(gam\right) \end{gathered}$$

-          Khi cho Y tác dụng lần lượt với các dung dịch BaCl₂ và AgNO₃ thì thu được kết tủa gồm:$\left\{\begin{array}{l}{n}_{BaS{O}_4}=n_{S{O}_4^{2-}}=n_{B{a}^{2+}}=0,455mol\\ n_{AgCl}=2n_{BaC{l}_2}=0,91mol\end{array}\right.\Rightarrow n_{Ag}=n_{F{e}^{2+}}=\frac{m\downarrow -233n_{BaS{O}_4}-143,5n_{AgCl}}{108}=0,18\left(mol\right)$

-          Khí Z chứa các khí H₂ (0,02 mol), CO₂ (0,11 mol), NO (0,04 mol).

$\stackrel{BT:N}{\to }{n}_{Cu{\left(N{O}_3\right)}_2}=0,5(n_{NO}+n_{N{H}_4^{+}}-n_{NaN{O}_3})=0,01mol.$

Từ (1) $\to 3n_{F{e}^{3+}}+2n_{M{g}^{2+}}=0,46\left(2\right)$

Và $24n_{M{g}^{2+}}+56n_{F{e}^{3+}}=6,72\left(3\right).$  Từ (2), (3) ta suy ra: $\left\{\begin{array}{l}{n}_{F{e}^{3+}}:0,06\\ n_{M{g}^{2+}}:0,14\end{array}\right.\left(mol\right)$

$\Rightarrow n_{Fe\left(X\right)}=(n_{F{e}^{3+}}+n_{F{e}^{2+}})-\underset{n_{C{O}_2}}{\underbrace{n_{FeC{O}_3}}}=0,06+0,18-0,11=0,13\left(mol\right).$

Vậy $\%m_{Fe}=\frac{0,13.56}{27,2}.100\%=26,76\%.$

Chọn đáp án B.