Hòa tan hoàn toàn 15,2 gam hỗn hợp X gồm FeO (0,02 mol)

* Hướng dẫn giải

Đáp án D

Đặt x, y, z là số mol NO; NO₂;CO₂

$$\begin{gathered} m_Y=30x+46y+44z=\frac{2.64\left(x+y+z\right)}{3}\left(1\right) \\ {n}_{Fe{\left(N{O}_3\right)}_2}=\frac{x+y}{2};n_{FeC{O}_3}=z \\ {m}_X=0,02.72+\frac{180\left(x+y\right)}{2}+116z+a=15,2\left(2\right) \\ {n}_{HCl}=4a+2y+2z+0,02.2 \end{gathered}$$

$\to$m muối $=0,02.56+\frac{56\left(x+y\right)}{2}+56z+a+35,5\left(4x+2y+2z+0,02.2\right)=18,88\left(3\right)$

Bảo toàn electron

$$\begin{gathered} 0,02+\frac{x+y}{2}+z+\frac{2a}{64}=3x+y+n_{Ag} \\ \to n_{Ag}=\frac{a}{32}-2,5x-0,5y+z+0,02 \\ m\downarrow =143,5\left(4x+2y+2z+0,02.2\right)+108\left(\frac{a}{32}-2,5x-0,5y+z+0,02\right)=50,24\left(4\right) \\ \left(1\right)\left(2\right)\left(3\right)\left(4\right)\to x=0,02;y=0,06;z=0,04;a=1,92 \\ m=0,02.56+\frac{56\left(x+y\right)}{2}+56z+a+17\left(4x+2y+2z+0,02.2\right)=12,96 \\ \to m+a=14,88 \end{gathered}$$