Đáp án đúng: C
Giải chi tiết:BTNT “N”: nNaNO3 = nNO + nNO2 = 0,02 + 0,1 = 0,12 (mol)
\(15,6(g)\left\{ \matrix{ MgO \hfill \cr CuO \hfill \cr F{e_3}{O_4} \hfill \cr Cu \hfill \cr Fe \hfill \cr} \right. + 200(g)\,{\rm{dd}}\left\{ \matrix{ NaN{O_3}:0,12 \hfill \cr {H_2}S{O_4}: \hfill \cr} \right. \to \left| \matrix{ {\rm{dd}}\,X\left\{ \matrix{ M{g^{2 + }} \hfill \cr C{u^{2 + }} \hfill \cr F{e^{2 + }} \hfill \cr F{e^{3 + }} \hfill \cr N{a^ + } \hfill \cr S{O_4}^{2 - } \hfill \cr} \right.\buildrel { + Ba{{(OH)}_2}vd} \over \longrightarrow Y\underbrace {\left\{ \matrix{ Mg{(OH)_2} \hfill \cr Cu{(OH)_2} \hfill \cr Fe{(OH)_2} \hfill \cr Fe{(OH)_3} \hfill \cr BaS{O_4} \hfill \cr} \right.}_{98,63(g)}\buildrel {{t^o}} \over \longrightarrow Z\underbrace {\left\{ \matrix{ MgO \hfill \cr CuO \hfill \cr F{e_2}{O_3} \hfill \cr BaS{O_4} \hfill \cr} \right.}_{93,93(g)} \hfill \cr \hfill \cr NO:0,02 \hfill \cr N{O_2}:0,1 \hfill \cr} \right.\)
Đặt nO(hh) = x mol
=> mMg, Cu, Fe = 15,6 – 16x (g)
Do H2SO4 pư vừa đủ nên ta có: nH+ = 4nNO + 2nNO2 + 2nO(hh) = 4.0,02 + 2.0,1 + 2x = 2x+0,28 (mol)
=> nH2SO4 = x+0,14 (mol)
nOH-(Y) = 2nMg2+ + 2nCu2+ + 2nFe2+ + 3nFe3+ = 2nSO42- - nNa+ = 2(x+0,14)-0,12 = 2x+0,16 (mol)
mY = mMg,Cu,Fe + mOH-(Y) + mBaSO4 => 15,6-16x+17(2x+0,16)+233(x+0,14) = 98,63 => x = 0,19
=> mMg, Cu, Fe =15,6 – 16x = 12,56 (g); nBaSO4 = x+0,14 = 0,33 mol; nOH-(Y) = 2x+0,16 = 0,54 mol
Ta có: mO(Z) = mZ – mMg,Cu,Fe – mBaSO4 = 93,93-12,56-0,33.233 = 4,48 gam
=> nO(Z) = 0,28 mol
\(Y\left\{ \matrix{ Mg{(OH)_2}:a \hfill \cr Cu{(OH)_2}:b \hfill \cr Fe{(OH)_2}:c \hfill \cr Fe{(OH)_3}:d \hfill \cr BaS{O_4}:0,33 \hfill \cr} \right.\buildrel {{t^o}} \over \longrightarrow Z\left\{ \matrix{ MgO:a \hfill \cr CuO:b \hfill \cr F{e_2}{O_3}:0,5c + 0,5d \hfill \cr BaS{O_4}:0,33 \hfill \cr} \right.\)
nOH-(Y) = 2a + 2b + 2c + 3d = 0,54 (1)
nO(Z) = a + b + 1,5c + 1,5d = 0,28 (2)
Lấy 2(2)-(1) được: c = 0,02 mol
=> nFeSO4 = c = 0,02 mol
BTKL: m dd sau pư = 15,6 + 200 – 0,02.30 – 0,1.46 = 210,4 gam
=> C%FeSO4 = (0,02.152/210,4).100% = 1,445% gần nhất với 1,45%
Đáp án C
