15g VO react with 15.7 g iron (III) oxide according to: 2VO+3Fe2O3->V2O5+6FeO How many grams of V2O5 are produced? limiting and excess reagent
First we should examine if the equation is balanced.
`2VO + 3Fe2O3 -> V2O5 + 6FeO` -> the equation is balanced.
Next, since the two reactants are given, we should look for the limiting reagent.
`15 grams VO *(1 mol e VO)/(66.94 grams VO) * (1 mol e V2O5)/(2 mol es VO)`
= 0.1120 moles V2O5
`15.7 grams Fe2O3 *(1 mol e Fe2O3)/(159.687 grams Fe2O3) * (1 mol e V2O5)/(3 mol es Fe2O3)`
= 0.03277 moles V2O5
We are using the limiting reagent method to determine which reactant will produce the smaller amount of product. Looking at our answers, we can see that Fe2O3 produces the smaller amount of V2O5 therefore, Fe2O3 is the limiting reagent. Apparently, we will use the value of V2O5 derived from Fe2O3 (we disregard the value we got from VO).
Finally we solve for the mass of V2O5:
`0.03277 mol es V2O5 * (181.88 grams V2O5)/(1 mol e V2O5)`
= 5.96 grams V2O5