Đáp án đúng: C
73.
$\underbrace{Fe,\,F{{e}_{3}}{{O}_{4}},\,Fe{{(N{{O}_{3}})}_{2}}}_{m\,(g)\,X}+\underbrace{KHS{{O}_{4}}}_{0,32\,mol}\to \left\langle \begin{array}{l}\overbrace{NO}^{0,04\ mol}+{{H}_{2}}O\\\underbrace{\overbrace{{{K}^{+}}}^{0,32\,mol},\,F{{e}^{2+}},\,F{{e}^{3+}},N{{O}_{3}}^{-},\overbrace{S{{O}_{4}}^{2-}}^{0,32\,mol}}_{59,04\,(g)\ \text{dd Y}}\end{array} \right.$$\xrightarrow{+\ NaOH}Fe{{(OH)}_{2}},\,Fe{{(OH)}_{3}}$
$\xrightarrow{BT:\,H}{{n}_{{{H}_{2}}O}}=\frac{{{n}_{KHS{{O}_{4}}}}}{2}=0,16\ mol$$\xrightarrow{BTKL}{{m}_{X}}={{m}_{Y}}+30{{n}_{NO}}+18{{n}_{{{H}_{2}}O}}-136{{n}_{KHS{{O}_{4}}}}=19,6\,(g)$
– Ta có:$2{{n}_{F{{e}^{2+}}}}+3{{n}_{F{{e}^{3+}}}}={{n}_{NaOH}}=0,44\ mol$$\xrightarrow{BTDT(Y)}{{n}_{N{{O}_{3}}^{-}}}={{n}_{{{K}^{+}}}}+2{{n}_{F{{e}^{2+}}}}+3{{n}_{F{{e}^{3+}}}}-2{{n}_{S{{O}_{4}}^{2-}}}=0,12\ mol$
$\xrightarrow{BT:\,N}{{n}_{Fe{{(N{{O}_{3}})}_{2}}}}=\frac{{{n}_{NO}}+{{n}_{N{{O}_{3}}^{-}}}}{2}=0,08\ mol$
⇒%mFe(NO3)2 = 73,46.
