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7.2g of an oxide of iron react with excess carbon monoxide to give 5.0g of
iron.what is the empirical formula of this oxide?
chemistry_MC(FORM 4)steps
2008-05-11 10:52 pm
回答 (1)
2016-03-09 1:47 am
FeXOY+YCO-->XFe+YCO2
Molar mass of iron: 55.8
Molar mass of Oxygen: 16
The number of mole of FeXOY: 7.2/(X*55.8+Y*16)
The number of mole of Fe form=5/55.8=0.0896
7.2/(X*55.8+Y*16)=0.0896/X
7.2X=0.0896(X*55.8+Y*16)
7.2X=5X+1.433Y
2.2X=1.433Y
X=0.652Y
X:Y=2:3
So the answer is Fe2O3
Ans check
LHS=7.2/(55.8*2+16*3)
=0.0451
RHS=0.0896/2
=0.0448
Molar mass of iron: 55.8
Molar mass of Oxygen: 16
The number of mole of FeXOY: 7.2/(X*55.8+Y*16)
The number of mole of Fe form=5/55.8=0.0896
7.2/(X*55.8+Y*16)=0.0896/X
7.2X=0.0896(X*55.8+Y*16)
7.2X=5X+1.433Y
2.2X=1.433Y
X=0.652Y
X:Y=2:3
So the answer is Fe2O3
Ans check
LHS=7.2/(55.8*2+16*3)
=0.0451
RHS=0.0896/2
=0.0448
收錄日期: 2021-04-16 14:00:56
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080511000051KK01386