Đốt m gam hỗn hợp E gồm Al, Fe và

Chọn đáp án B

Giải thích:

$\begin{array}{l}+m_Y=m_X+m_{CO}-n_{(CO,C{O}_2)}=\mathrm{34,4}+\mathrm{0,3.28}-\mathrm{0,3.36}=32gam.\\ +BTKL(Y+HN{O}_3)\Rightarrow n_{H_{2}O}=\frac{32+\mathrm{1,7.63}-\mathrm{117,46}-\mathrm{0,2.16,75.2}}{18}=\mathrm{0,83}\Rightarrow n_{N{H_4}^{+}}=\frac{\mathrm{1,7}-\mathrm{0,83.2}}{4}=\mathrm{0,01.}\\ +\left\{\begin{array}{l}{n}_{NO}+n_{N_{2}O}=\mathrm{0,2}\\ 30n_{NO}+44n_{N_{2}O}=\mathrm{6,7}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{NO}=\mathrm{0,15}\\ n_{N_{2}O}=\mathrm{0,05}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{N{O_3}^{-}taomuoi}=\mathrm{1,7}-\mathrm{0,01}-\mathrm{0,15}-\mathrm{0,05.2}=\mathrm{1,44}\\ m_{(Al,Fe,Cu)}=\mathrm{117,46}-\mathrm{1,44.62}-\mathrm{0,01.18}=28gam.\end{array}\right.\end{array}$