Can anyone answer this? chem prob with unknown element?
What theoretical weight of the oxide is formed when 28g is heated to excess oxygen? what is the percent yield, if 38g of the oxide is produced?
Answer:
As I told you
2.4 / 16 = 0.15 moles O
2 :3 = x : 0.15
x = 0.1 Moles A
8 - 2.4 = 5.6 g A
1 mole : x grams = 5.6 : x
x = 56 g/mol = atomic weight of A
the formula is A2O3
molar mass = 160 g/mol
And now
The reaction between A and O2 is
4 A +3 O2 >> 2A2O3
moles of A = 28 / 56 = 0.5
the ratio between A and A2O3 is 4 : 2
4 : 2 = 0.5 : x
x = 0.25 moles A2O3 produced
0.25 x 160 g/mol = 40 g A2O3 produced
if we get only 38 g
38 : 40 = x : 100
x = 95 %
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