Question

If 143 grams of chromium react with an excess of oxygen, as shown in the balanced chemical equation below, how many grams of chromium oxide can be formed? Please show all your work for the calculations for full credit.

4Cr + 3O2 yields 2Cr2O3

Answer 1

4 Cr + 3 O2 → 2 Cr2O3

(143 g Cr) / (51.99616 g Cr/mol) x (2 mol Cr2O3 / 4 mol Cr) x
(151.99061 g Cr2O3/mol) = 209 g Cr2O3

Answer 2

Answer: 209 g

Explanation:

`4Cr+3O_2\rightarrow 2Cr_2O_3`

Molecular weight of Cr= 52g/mol

Molecular weight of
`Cr_2O_3`= 152g/mol

As Oxygen is present in excess, Chromium is the limiting reagent as it limits the formation of product.

`4* 52g=208g` of Chromium produces
`2* 152g=304g` of
`Cr_2O_3`

Thus 143 g of Chromium(Cr) produces=
`(304)/(208)* 143=209g` of Chromium oxide(
`Cr_2O_3`)