A compound of the 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,

Question

a. What is the atomic weight of A?
b. What is the weight of one mole of the oxide?
c. What theoretical weight of the oxide is formed when 28 grams of A is heated in excess oxygen? What is the % yield, if 38 grams of the oxide was produced?

Answer 1

A2O3 is the compound. If 8.0 g of A2O3 contains 2.4 g oxygen, then mass of A must be 8.0-2.4 = 5.6. If we were trying to determine the formula we would convert grams to moles and find the ratio of the moles to each other. In this case, we already have the mole ratio; we want to determine the molar mass.
moles A = 5.6g/molar mass of x
moles O = 2.4/15 = 0.15
If we want the ratio to be 2:3 and we know O is 0.15 mole, what is the moles A? That must be 0.667*0.15 = 0.1.
Then moles = 5.6/x
0.1 = 5.6/x
x = 56 for the atomic mass.

mass of 1 mole is the molar mass. 2*atomic mass Z + 3*atomic mass O = molar mass.

For c part.
Write and balance the equation between A and O (but you can use Ba for A if you wish).
Convert 28.0 g Ba to moles. moles = grams/molar mass.
Using the coefficients in the balanced equation, convert moles Ba to moles of the oxide.
The g oxide = moles x molar mass. This is the theoretical yield.

Finally, percent yield = (actual yield/theoretical yield)*100 = ?

Answer 2

thanks for the help :)