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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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