Hòa tan hết 30 gam rắn gồm Mg, MgO, MgCO3 trong HNO3 thấy có 2,15 mol HNO3 phản ứng. Sau phản ứng thu được 4,48

* Hướng dẫn giải

$\begin{array}{l}+\left\{\begin{array}{l}{n}_{NO}+n_{C{O}_2}=0,2\\ 30n_{NO}+44n_{C{O}_2}=0,2.18,5.2=7,4\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{NO}=0,1\\ n_{C{O}_2}=0,1\end{array}\right.\Rightarrow n_{MgC{O}_3}=0,1\\ +Sơ\text{\hspace{0.17em}\hspace{0.17em}đồ\hspace{0.17em}\hspace{0.17em}}phản\text{\hspace{0.17em}\hspace{0.17em}ứ}ng:\\ \underset{hỗn\text{\hspace{0.17em}h\hspace{0.17em}ợ}pX}{\underbrace{\left\{\begin{array}{l}Mg,MgO\\ MgC{O}_3\end{array}\right\}}}+\underset{2,15mol}{\underbrace{HN{O}_3}}\stackrel{}{\to }\underset{ddY}{\underbrace{\left\{\begin{array}{l}M{g}^{2+};N{H_4}^{+}\\ N{O_3}^{-}\end{array}\right\}}}+\left\{\begin{array}{l}NO\\ C{O}_2\end{array}\right\}\end{array}$

$\begin{array}{l}+\left\{\begin{array}{l}{n}_{Mg}=a\\ n_{MgO}=b\end{array}\right.;\left\{\begin{array}{l}{n}_{N{H_4}^{+}}=x\\ BTN:n_{N{O_3}^{-}trongY}=2,05-x\end{array}\right.\\ +\left\{\begin{array}{l}BTE:2a=8x+0,1.3\\ BTĐTtrongY:2(a+b+0,1)+x=2,05-x\\ m_X=24a+40b+0,1.84=30\end{array}\right.\Rightarrow \left\{\begin{array}{l}a=0,65\\ b=0,15\\ x=0,125\end{array}\right.\\ \Rightarrow m_{muốitrongY}=\boxed{143,2gam}\end{array}$

Chọn B