# What theoretical weight of the oxide is formed when 28 grams of A is heated in excess oxygen and what is the % yield if 38 grams of the oxide is produced? Given: A compound of element A and oxygen...

What theoretical weight of the oxide is formed when 28 grams of A is heated in excess oxygen and what is the % yield if 38 grams of the oxide is produced?

Given: A compound of element A and oxygen has a mole ratio of A:O = 2:3. If 8.0 grams of the oxide contains 2.4 grams of the oxygen.

jerichorayel | Certified Educator

The reaction would be:

`2 A + 3 O -> A_2O_3`

Moles of O in the oxide:

`= (2.4 grams)/(16.0 (grams)/(mol e)) = 0.15 mol es`

Since the ratio is 2:3 then:

`2:3 and?:0.15`

`= 0.15 * 2/3 = 0.1 mol es A`

Molar mass of A:

`molar mass A= (5.6)/(0.1)`

`molar mass A = 56 (g)/(mol e)`

Molar Mass of A2O3

`= 2(56) + 3(16)`

`= 160(g)/(mol e)`

For the theoretical and percent yield:

`28 grams of A * (1 mol e A)/(56 grams A) * (1 mol e A_2O_3)/(2 mol es A) * (160 grams)/(1 mol e A)`

`= 40 grams A_2O_3` -> theoretical yield.

`Percent yield = 38/40 * 100`

`Percent yield = 95 %`