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The rate of flow in stream with 40% O2 and 60% CO2 is 10 mole/hr, the contained was initially filled with 6 mole of O2, what will be the percentage of O2 in product stream?

The rate of flow in stream with 40% O2 and 60% CO2 is 10 mole/hr, the contained was initially filled with 6 mole of O2, what will be the percentage of O2 in product stream? Correct Answer

Percentage of O2 = 10/(10 + 6)*100 = 62.5%.

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