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What theoretical weight of the oxide is formed when 28 grams of A is heated in excess...
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.
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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:
`= 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 %`
Posted by jerichorayel on April 25, 2013 at 5:19 PM (Answer #1)
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