Cho m gam hỗn hợp bột X gồm FexOy

Chọn đáp án B

Giải thích:

$\begin{array}{l}+\mathrm{3,2}gamkimloailaCu.\\ +\left\{\begin{array}{l}F{e}_x{O}_y\\ Cu,CuO\end{array}\right\}\stackrel{\mathrm{0,3}molHCl}{\to }\left\{\begin{array}{l}F{e}^{2+}\\ C{u}^{2+}\\ C{l}^{-}:\mathrm{0,3}\end{array}\right\}\stackrel{AgN{O}_{3}du}{\to }\left\{\begin{array}{l}AgCl:\mathrm{0,3}mol\\ Ag:\frac{\mathrm{51,15}-\mathrm{0,3.143,5}}{108}=\mathrm{0,075}mol\end{array}\right\}\\ \Rightarrow n_{F{e}^{2+}}=n_{A{g}^{+}pu}=n_{Ag}=\mathrm{0,075}mol\Rightarrow n_{C{u}^{2+}}=(\mathrm{0,3}-\mathrm{0,075.2}):2=\mathrm{0,075}mol.\\ +2n_{O^{2-}}=n_{C{l}^{-}}=\mathrm{0,3}mol\Rightarrow n_{O^{2-}}=\mathrm{0,15}mol.\\ \Rightarrow m_X=\mathrm{3,2}+\mathrm{0,075.56}+\mathrm{0,075.64}+\mathrm{0,15.16}=\mathrm{14,6}gannhatvoi\mathrm{14,5.}\end{array}$