Hòa tan hết 0,6 mol hỗn hợp X gồm Mg, Fe(NO3)3 và Fe(NO3)2 trong...

Bài toán: $\underbrace{X}_{0,6mol}\left\{ \begin{aligned}
& Mg \\
& Fe{{\left( N{{O}_{3}} \right)}_{3}} \\
& Fe{{\left( N{{O}_{3}} \right)}_{2}} \\
\end{aligned} \right.\xrightarrow[0,18\left( mol \right)]{HN{{O}_{3}}}\left\langle \begin{aligned}
& \underbrace{Z}_{0,1mol}\left\{ \begin{aligned}
& NO \\
& {{N}_{2}}O \\
\end{aligned} \right. \\
& \underbrace{muo\acute{a}i Y}_{103,3\left( gam \right)}\left\langle \begin{aligned}
& \xrightarrow{t{}^\circ }\underbrace{Ra\acute{e}n}_{31,6\left( gam \right)}\left\{ \begin{aligned}
& MgO \\
& F{{e}_{2}}{{O}_{3}} \\
\end{aligned} \right. \\
& \xrightarrow{NaOH}Ke\acute{a}t tu\hat{u}a:42,75\left( gam \right) \\
\end{aligned} \right. \\
\end{aligned} \right.$
$X\left\{ \begin{aligned}
& Mg:a \\
& Fe:b \\
& NO_{3}^{-}:c \\
\end{aligned} \right.\to a+b=0,6(*)$
$\left\{ \begin{aligned}
& BTNT\left( Mg \right):{{n}_{MgO}}=b \\
& BTNT\left( Fe \right):{{n}_{F{{e}_{2}}{{O}_{3}}}}=\dfrac{b}{2} \\
\end{aligned} \right.\to 40a+80b=31,6(**)\xrightarrow{(*)}\left\{ \begin{aligned}
& a=0,41 \\
& b=0,19 \\
\end{aligned} \right.$
Hiđroxit: $\left\{ \begin{aligned}
& {{M}^{n+}} \\
& O{{H}^{-}} \\
\end{aligned} \right.\to {{n}_{O{{H}^{-}}}}=\dfrac{42,75-0,41.24-0,19.56}{17}=1,31\left( mol \right)$
$Y\left\{ \begin{aligned}
& {{M}^{n+}};NO_{3}^{-} \\
& N{{H}_{4}}N{{O}_{3}} \\
\end{aligned} \right.\to {{n}_{N{{H}_{4}}N{{O}_{3}}}}=\dfrac{103,3-1,31.62-0,41.24-0,19.56}{80}=0,02\left( mol \right)$
BTNT(H): ${{n}_{{{H}_{2}}O}}=\dfrac{1,08-0,24}{2}=0,5\left( mol \right)$
BTNT(O): 3c = 0,5 + 0,1 + 1,31.3 + 0,02.3 - 1,08.3 = 1,35 → c = 0,45
$\left\{ \begin{aligned}
& Fe{{\left( N{{O}_{3}} \right)}_{3}}:x \\
& Fe{{\left( N{{O}_{3}} \right)}_{2}}:y \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& x+y=0,19 \\
& 9x+6y=0,45 \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& x=0,12 \\
& y=0,07 \\
\end{aligned} \right.\to \%{{m}_{Fe{{\left( N{{O}_{3}} \right)}_{3}}}}=35,01\%$
→ Chọn đáp án B.