Em tham khảo nha:
Câu c đề sai rồi em nhé
\(\begin{array}{l}
2)\\
a)\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
b)\\
n{H_2} = \dfrac{V}{{22,4}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
nF{e_2}{O_3} = \dfrac{{0,3}}{3} = 0,1\,mol\\
x = mF{e_2}{O_3} = n \times M = 0,1 \times 160 = 16g\\
nFe = 2nF{e_2}{O_3} = 0,2\,mol\\
y = mFe = n \times M = 56 \times 0,2 = 11,2g\\
c)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
n{H_2} = nFe = 0,2\,mol\\
n{H_2} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol
\end{array}\)
