Gọi CTTQ: FexOy
Pt: 2xFe + yO2 --to--> 2FexOy
.................0,025 mol-> \(\dfrac{0,05}{y}\) mol
Áp dụng ĐLBTKL, ta có:
mFe + mO2 = mFexOy
=> mO2 = mFexOy - mFe = 2,9 - 2,1 = 0,8 (g)
=> nO2 = \(\dfrac{0,8}{32}=0,025\) mol
Ta có: 2,9 = \(\dfrac{0,05}{y}\left(56x+16y\right)\)
\(\Leftrightarrow2,9=\dfrac{2,8x}{y}+0,8\)
\(\Leftrightarrow\dfrac{2,8x}{y}=2,1\)
\(\Leftrightarrow2,8x=2,1y\)
\(\Leftrightarrow\dfrac{x}{y}=\dfrac{2,1}{2,8}=\dfrac{3}{4}\)
Vậy CTHH của oxit sắt: Fe3O4
P/s: bn có thể lm cách khác
nFe = 0,0375 mol
nFexOy = \(\dfrac{2,9}{56x+16y}\) mol
2xFe + yO2 → 2FexOy
0,0375...............\(\dfrac{0,0375}{x}\)
⇒ \(\dfrac{2,9}{56x+16y}\)= \(\dfrac{0,0375}{x}\)
⇔ 2,9x = 2,1x + 0,6y
⇔ 0,8x = 0,6y
⇔ \(\dfrac{x}{y}\)= \(\dfrac{0,6}{0,8}\)= \(\dfrac{3}{4}\)
⇒ CTHH : Fe3O4