Cho 33,4 gam hỗn hợp rắn X gồm Mg; MgO; Fe(NO3)2 và

Đáp án C.

$39,4X\left\{\begin{array}{l}\text{Mg:\hspace{0.17em}\hspace{0.17em}x}\\ \text{MgO:\hspace{0.17em}\hspace{0.17em}y}\\ \text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{:\hspace{0.17em}\hspace{0.17em}z}\\ {\text{FeCO}}_{\text{3}}\text{:0,1\hspace{0.17em}\hspace{0.17em}}\left({\text{n}}_{{\text{CO}}_{\text{2}}}\right)\end{array}\right.\underset{{\text{HNO}}_{\text{3}}:0,166}{\overset{\text{HCl:\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1,29}{\to }}\left\{\begin{array}{l}0,163\text{\hspace{0.17em}\hspace{0.17em}Z}\left\{\begin{array}{l}{\text{N}}_{\text{2}}\text{O}\\ {\text{N}}_{\text{2}}\\ {\text{CO}}_{\text{2}}\text{:0,1}\end{array}\right.\\ Y\stackrel{{\text{AgNO}}_3}{\to }191,595\left\{\begin{array}{l}\text{Ag}\\ \text{AgCl:\hspace{0.17em}1,29\hspace{0.17em}}\left({\text{n}}_{\text{HCl}}\right)\end{array}\right.\end{array}\right.$

$\to n_{Ag}=\frac{191,595-1,29.\left(108+35,5\right)}{108}=0,06\to n_{F{e}^{2+}\left(Y\right)}=0,06$

$Y\left\{\begin{array}{l}{\text{Fe}}^{\text{2+}}\text{:\hspace{0.17em}\hspace{0.17em}0,06}\\ {\text{Fe}}^{\text{3+}}{\text{:\hspace{0.17em}\hspace{0.17em}BT\hspace{0.17em}\hspace{0.17em}Fe:\hspace{0.17em}n}}_{{\text{Fe}}^{\text{3+}}}\text{=}\left(\text{z+0,1-0,06}\right)\text{=z+0,04}\\ {\text{Mg}}^{\text{2+}}\text{:\hspace{0.17em}\hspace{0.17em}x+y}\\ {\text{Cl}}^{\text{-}}\text{:1\hspace{0.17em}\hspace{0.17em},29}\\ {\text{NO}}_{\text{3}}^{\text{-}}{\text{:\hspace{0.17em}\hspace{0.17em}n}}_{{\text{NO}}_{\text{3}}^{\text{-}}}{\text{+n}}_{{\text{Cl}}^{\text{-}}}{\text{=n}}_{\text{ion+}\left(\text{Y}\right)}{\text{=n}}_{\text{KOH}}\text{=1,39}\to {\text{n}}_{{\text{NO}}_{\text{3}}^{\text{-}}}=0,1\\ {\text{NH}}_{\text{4}}^{\text{+}}{\text{:\hspace{0.17em}\hspace{0.17em}BT\hspace{0.17em}\hspace{0.17em}N:n}}_{{\text{NH}}_{\text{4}}^{\text{+}}}={\text{2n}}_{\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}{\text{+n}}_{{\text{HNO}}_{\text{3}}}{\text{-2n}}_{{\text{N}}_{\text{2}}{\text{O,N}}_{\text{2}}}{\text{-n}}_{{\text{NO}}_{\text{3}}^{\text{-}}}=\left(2\text{z}+0,166-0,063.2-0,1\right)=2\text{z}-0,06\end{array}\right.$

$\to \left\{\begin{array}{l}y+6\text{z}+0,1.3=0,68\left(n_{O\left(X\right)}\right)\\ 0,06.2+3.\left(\text{z}+0,04\right)+2.\left(x+y\right)+\left(2\text{z}-0,06\right)=1,39\left(n_{KOH}\right)\\ 24x+40y+180\text{z}+11,6=33,4\end{array}\right.\to \left\{\begin{array}{l}x=0,4\\ y=0,08\\ \text{z}=0,05\end{array}\right.$

BT H: $n_{H_{2}O}=0,648$

BT O: $n_{N_{2}O}=0,03$

$\to n_{N_2}=0,033$