1.

What is mass of `O_(2)` required to produce `960gm` of `O_(3)` if `%` yield of reaction `30_(2)rarr 2O_(3)` is `50%` ? .

Answer» `3O_(2)rarr 2O_(3)`
`(960)/(48) = 20mol` = Actual yield
Let `X` mol `O_(2)` be required
Theoretical yield of `O_(3)((2)/(3)X)rArr50((2)/(2/(3)X))(100)`
`X = 60 mol = 60 xx 32 = 1920 g O_(2)` required .


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