Hoà tan hoàn toàn m g hỗn hợp X gồm Mg, Fe, FeCO3, Cu(NO3)2 vào dung dịch chứa H2SO4 và 0,045 mol NaNO3, thu được dung

* Hướng dẫn giải

Đáp án B

\(\underbrace {Mg,Fe,FeC{O_3},Cu{{(N{O_3})}_2}}_{m(g)\,\,X} + \underbrace {{H_2}S{O_4},\overbrace {NaN{O_3}}^{0,045\,\,mol}}_{dung\,\,dich\,\,hon\,\,hop} \to \underbrace {M{g^{2 + }},F{e^{a + }},C{u^{2 + }},\overbrace {N{a^ + }}^{0,045\,\,mol},NH_4^ + ,SO_4^{2 – }}_{62,605(g)\,\,Y} + \underbrace {\overbrace {{H_2}}^{0,02\,\,mol},C{O_2},NO}_{0,17\,\,mol\,\,hon\,\,hop\,\,Z}\)

Cho  \(\underbrace {M{g^{2 + }},F{e^{a + }},C{u^{2 + }},NH_4^ + ,SO_4^{2 – }}_{62,605(g)\,Y} \to \underbrace {Fe{{(OH)}_a},Cu{{(OH)}_2},Mg{{(OH)}_2}}_{31,72(g) \downarrow } + N{a_2}S{O_4}\)

\( \Rightarrow a{n_{F{e^{a + }}}} + 2{n_{M{g^{2 + }}}} + 2{n_{C{u^{2 + }}}} + {n_{NH_4^ + }} = {n_{NaOH}} = 0,865\,\,\,\,\,\,\,\,\left( 1 \right)\)

\(\begin{array}{l}
\to {n_{{H_2}S{O_4}}} = {n_{SO_4^{2 – }}} = 0,455\,\,mol\\
\Rightarrow {m_{ \downarrow \max }} = 56{n_{F{e^{a + }}}} + 24{n_{M{g^{2 + }}}} + 64{n_{C{u^{2 + }}}} + 17\left( {{n_{O{H^ – }}} – {n_{NH_4^ + }}} \right) \to 56{n_{F{e^{a + }}}} + 24{n_{M{g^{2 + }}}} + 64{n_{C{u^{2 + }}}} = 17,015 + 17{n_{NH_4^ + }}
\end{array}\)

Ta có: \({m_Y} = 56{n_{F{e^{a + }}}} + 24{n_{M{g^{2 + }}}} + 64{n_{C{u^{2 + }}}} + 23{n_{N{a^ + }}} + 18{n_{NH_4^ + }} + 96{n_{SO_4^{2 – }}}\)

\(\begin{array}{l}
\to 62,605 = 17,015 + 17{n_{NH_4^ + }} + 23.0,045 + 18{n_{NH_4^ + }} + 96.0,455 \to {n_{NH_4^ + }} = 0,025(mol)\\
\to {n_{{H_2}O}} = \frac{{2{n_{{H_2}S{O_4}}} – 4{n_{NH_4^ + }} – 2{n_{{H_2}}}}}{2} = 0,385\,\,mol\\
\to {m_X} = {m_Y} + {m_Z} + 18{n_{{H_2}O}} – 85{n_{NaN{O_3}}} – 98{n_{{H_2}S{O_4}}} = 27,2(gam)
\end{array}\)

Khi cho Y tác dụng lần lượt với các dung dịch BaCl2 và AgNO3 thì thu được kết tủa gồm:

\(\left\{ \begin{array}{l}
{n_{BaS{O_4}}} = {n_{SO_4^{2 – }}} = {n_{B{a^{2 + }}}} = 0,455\,\,mol\\
{n_{AgCl}} = 2{n_{BaC{l_2}}} = 0,91\,\,mol
\end{array} \right. \Rightarrow {n_{Ag}} = {n_{F{e^{2 + }}}} = \frac{{m \downarrow – 233{n_{BaS{O_4}}} – 143,5{n_{AgCl}}}}{{108}} = 0,18\,(mol)\)

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

\( \to {n_{Cu{{(N{O_3})}_2}}} = 0,5({n_{NO}} + {n_{NH_4^ + }} – {n_{NaN{O_3}}}) = 0,01\,\,mol.\)

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

Và \(24{n_{M{g^{2 + }}}} + 56{n_{F{e^{3 + }}}} = 6,72\,\,\,\,\left( 3 \right).\) 

Từ (2), (3) ta suy ra: \(\left\{ \begin{array}{l}
{n_{F{e^{3 + }}}}:0,06\\
{n_{M{g^{2 + }}}}:0,14
\end{array} \right.\,\,\,(mol)\)

\( \Rightarrow {n_{Fe(X)}} = ({n_{F{e^{3 + }}}} + {n_{F{e^{2 + }}}}) – \underbrace {{n_{FeC{O_3}}}}_{{n_{C{O_2}}}} = 0,06 + 0,18 – 0,11 = 0,13\,\,(mol).\)

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