\({n_{Fe}}\) = \(\dfrac{{35}}{56}\)= \(0,625(mol)\)
Phương trình hóa học :
\(3{H_2}\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,F{e_2}{O_3}\buildrel {{t^o}} \over\longrightarrow 2Fe + 3{H_2}O\)
3 mol 1 mol 2 mol
x mol \( \leftarrow \) 0,625 mol
\(x \)= \(\dfrac{{0,625 \times 3}}{{2}}\)= \(0,9375(mol)\)
\({V_{{H_2}}} = 0,9375 \times 22,4 = 21(l)\)
\(3CO\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,F{e_2}{O_3} \to 2Fe + \,\,\,\,3C{O_2}\)
3 mol 1 mol 2 mol
y mol \( \leftarrow \) 0,625 mol
\(y \)= \(\dfrac{{0,625 \times 3}}{{2}}\)= \(0,9375(mol)\)
\({V_{CO}} = 0,9375 \times 22,4 = 21(l)\).
=> Chọn D.
