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langtu_117
[TEX]a)\\R^0\rightarrow R^{n+}+ne\\N^{5+}+1e\rightarrow N^{4+}\\--------\\R^{0}\rightarrow R^{m+}+me\\2H^{+}+2e\rightarrow H_2\\nR=a(mol);nNO_2=3nH_2\\\rightarrow a.n=3.\frac{a.m}{2}\rightarrow n=\frac{3m}{2}\\\rightarrow \left\{\begin{matrix}n=3\\m=2\end{matrix}\right.[/TEX]
[TEX]\rightarrow R(NO_3)_3\ va \ RSO_4\\mRSO_4=\frac{62,81}{100}.mR(NO_3)_3\\ \rightarrow M_R+96=\frac{62,81}{100}(M_R+186)\rightarrow M_R=56\\\rightarrow R \ la \ Fe[/TEX]
[TEX]b)xFe+\frac{y}{2}O_2\rightarrow Fe_xO_y\\Fe + 6 HNO_3 \rightarrow Fe(NO_3)_3 + 3 H_2O + 3 NO_2\\nFe=a(mol)\rightarrow nO_2=\frac{a.\frac{y}{2}}{x}(mol);nNO_2=3a\\nO_2= \frac{22,22}{100}.nNO_2\rightarrow \frac{\frac{y}{2}}{x}=\frac{2}{3}\rightarrow \frac{x}{y}=\frac{3}{4}\rightarrow A:Fe_3O_4[/TEX]
[TEX]nA=0,09(mol)\\3Fe^{\frac{8}{3}+}\rightarrow 3Fe^{3+}+1e\\nN_xO_y=0,03(mol)\\n \ e \ nhuong=0,09(mol) \rightarrow n \ e \ nhan=0,09(mol)\\\rightarrow NO\\3 Fe_3O_4 + 28 HNO_3 \rightarrow 9 Fe(NO_3)_3 + 14 H_2O + NO\\nHNO_3 \ pu=\frac{28}{3}nFe_3O_4=0,84(mol)\rightarrow mHNO_3 \ can=66,15(g)[/TEX]
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(nhưng chắc tại mình ngu nhiều hơn ==")