Phương pháp giải:

1. Bảo toàn nguyên tố

2. Bảo toàn e

3. Bảo toàn điện tích

Lời giải chi tiết:

\(\begin{gathered}
+ \left\{ \begin{gathered}
{n_{NO}} + {n_{C{O_2}}} = 0,2 \hfill \\
30{n_{NO}} + 44{n_{C{O_2}}} = 0,2.18,5.2 = 7,4 \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
{n_{NO}} = 0,1 \hfill \\
{n_{C{O_2}}} = 0,1 \hfill \\
\end{gathered} \right. \to {n_{MgC{O_3}}} = 0,1 \hfill \\
+ So\,do\,phan\,ung: \hfill \\
\underbrace {\left\{ \begin{gathered}
Mg,MgO \hfill \\
MgC{O_3} \hfill \\
\end{gathered} \right\}}_{hon\,hop\,X} + \underbrace {HN{O_3}}_{2,15mol} \to \underbrace {\left\{ \begin{gathered}
M{g^{2 + }};N{H_4}^ + \hfill \\
N{O_3}^ - \hfill \\
\end{gathered} \right\}}_{{\text{dd}}\,Y}\left\{ \begin{gathered}
NO \hfill \\
C{O_2} \hfill \\
\end{gathered} \right\} \hfill \\
+ \left\{ \begin{gathered}
{n_{Mg}} = a \hfill \\
{n_{MgO}} = b \hfill \\
\end{gathered} \right.;\left\{ \begin{gathered}
{n_{N{H_4}^ + }} = x \hfill \\
BTNT\,N:{n_{N{O_3}^ - (Y)}} = 2,05 - x \hfill \\
\end{gathered} \right. \hfill \\
+ \left\{ \begin{gathered}
BTe:2a = 8x + 0,1.3 \hfill \\
BTDT\,trong\,Y:2(a + b + 0,1) + x = 2,05 - x \hfill \\
{m_X} = 24a + 40b + 0,1.84 = 30 \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
a = 0,65 \hfill \\
b = 0,15 \hfill \\
x = 0,125 \hfill \\
\end{gathered} \right. \hfill \\
\to {m_{muoi\,trong\,Y}} = 143,2(g) \hfill \\
\end{gathered} \)

Đáp án B