Calculate the weight of `FeO` produced from `2 g VO` and `5.75 g` of `Fe_(2)O_(3)`. Also report the limiting reagent.
Given : `VO+Fe_(2)O_(3) rarr FeO+V_(2)O_(5)`

Explanation

Balanced equation is
`{:(,2VO+,3Fe_(2)O_(3)rarr,6FeO+V_(2)O_(5),),("Molar before",2//67,5.75//160,,),("reaction",=0.0298,0.0359,,),("Mole after reaction","["0.0298-"("0.0359xx2")"//3"]",0,"["0.0359xx2"]""["0.0359xx1"]",):}`
`:.` Mole ratio in reaction is
`VO:Fe_(2)O_(3):FeO:V_(2)O_(5)::2:3:6:1`
`:.` Mole of `FeO` formed `=0.0359xx2`
`:.` Weight of `FeO` formed `= 0.0359xx2xx72`
`= 5.18 g`
The limiting reagent is the one which is totally used i.e., `Fe_(2)O_(3)` here.