In the reaction of canadium oxide (VO) with iron oxide `(Fe_2O_3)` the products are `V_2O_5` and `FeO`. How many grams of `V_2O_5` can be formed from 2.00 g of VO and 5.75 g of `Fe_2O_3`.

Explanation

`2VO+3Fe_2O_3toV_2O_5+6FeO`
Here `Fe_2O_3` is a limiting reagent.
`3 " mol of "Fe_2O_3-=1 " mol of "V_2O_5`
`(5.75)/(160)=0.036 " mol of "Fe_2O_3=(0.036)/(3)" mol of "V_2O_5`
grams `V_2O_5=(0.036)/(3)xx181.9=2.182g`