Hòa tan hết 32 gam hỗn hợp X gồm Fe

Chọn đáp án A

Giải thích:

$\begin{array}{l}+\left\{\begin{array}{l}F\text{e}:amol\\ O:bmol\end{array}\right\}\stackrel{quydoi}{\leftarrow }\underset{X,32gam}{\underbrace{\left\{\begin{array}{l}F\text{e}\\ F{\text{e}}_x{O}_y\end{array}\right\}}}\underset{\left(1\right)}{\overset{\mathrm{1,7}molHN{O}_3}{\to }}\underset{Y}{\underbrace{\left\{\begin{array}{l}F{\text{e}}^{3+}\\ N{O_3}^{-}\\ H^{+}\end{array}\right\}}}\underset{\left(2\right)}{\overset{\mathrm{0,2}molCu}{\to }}\underset{Z}{\underbrace{\left\{\begin{array}{l}F{\text{e}}^{2+}:amol\\ C{u}^{2+}:\mathrm{0,2}mol\\ N{O_3}^{-}\end{array}\right\}}}\\ +n_{NOo\left(1\right),\left(2\right)}=cmol.\\ +\left\{\begin{array}{l}BTEo\left(1\right),\left(2\right):2\text{a+0,2.2}=2b+3c\\ m_X=56\text{a}+16b=32\\ n_{H^{+}}=2b+4c=\mathrm{1,7}\end{array}\right.\Rightarrow \left\{\begin{array}{l}a=\mathrm{0,5}\\ b=\mathrm{0,25}\\ c=\mathrm{0,3}\end{array}\right.\Rightarrow V=\mathrm{6,72}lít.\end{array}$