A cpd of element A and oxygen has a mole ratio of A:O = 2:3. If 8 grams of the oxide contains 2.4 g of oxygen,

Then what is: (a) what is the atomic wt of A? (b) What is the wt of one mole of oxide? (c) What theoretical wt of the oxide is formed when 28 grams is heated in excess oxygen? What is the % yield, if 38 grams of the oxide is produced? Can you help me answer this problem. (This is based on the law of definite composition/proportions) Thnx!

Answer:
a)
2.4 grams of oxygen is contained by 8 grams of the compound. So, the mass of compound that will contain 48 grams oxygen = 8*48/2.4 = 160 grams

48 grams of oxygen corresponds to 3 gram atom of oxygen. So, if we consider the compound to be A2O3, we have 160 grams as the gram molecular weight. The rest of the thing is A, that is 2 gram moles of A weigh (160-48) 112 grams. Thus, the atomic weight of A is 112/2 = 56 grams.

b)
The weight of the oxide is 160 grams

c)
28 grams of the metal corresponds to 28/56 = 0.5 moles. Each mole of the compound contains 2 moles of the metal. So 0.5 moles of the metal will form 0.25 moles of the oxide.
The weight of 0.25 moles of the oxide = 40 grams

d)
38 grams was yielded. So the percent yield = 38/40 *100 = 95 %

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A cpd of element A and oxygen has a mole ratio of A:O = 2:3. If 8 grams of the oxide contains 2.4 g of oxygen,

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