Question

Answer 1

\(n_{NO}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Gọi số mol Fe là a (mol)

Fe⁰ - 3e --> Fe⁺³

a--->3a

O₂⁰ + 4e --> 2O^-2

0,3-->1,2

N⁺⁵ + 3e --> N⁺²

        0,6<--0,2

Bảo toàn e: 3a = 1,8 => a = 0,6 (mol)

=> m_Fe = 0,6.56 = 33,6 (g)

=> D

Answer 2

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{NO}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_O=0,3.2=0,6\left(mol\right)\\ 4.chất.rắn:\left\{{}\begin{matrix}0,6\left(mol\right)O\\\dfrac{m}{56}\left(mol\right)Fe\end{matrix}\right.\\ PT:Fe^0\rightarrow Fe^{3+}+3e\\ O^0+2e\rightarrow O^{2-}\\ N^{+5}+3e\rightarrow N^{+2}\\ BT.e:3.n_{Fe}=2.n_O+3.n_{NO}\\ \Leftrightarrow3.\dfrac{m}{56}=2.0,6+3.0.2\\ \Leftrightarrow m=33,6\left(g\right)\)

Answer 3

4 chất rắn là Fe dư Fe2O3 FeO Fe3O4 chăng?