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  • one year ago

In an experiment, 12.0dm3 of oxygen, measured under room conditions, is used to burn completely 0.10mol of propan-1-ol. What is the final volume of gas, measured under room conditions? A 7.20dm3 B 8.40dm3 C 16.8dm3 D 18.00dm3 ans is B. Pls explain how

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  1. anonymous
    • one year ago
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    1) you have to write the combustion reaction of the O2 plus propan-1-ol. If this is a "combustion" the products are CO2 and H2O. 2) balance the reaction to find the correspondent coefficients. 3) determine the limiting reactant. Calculate the moles of CO2 (g) 4) the tricky part is that O2 and the CO2 at room conditions (293K, 1atm) are gases and the propan-1-ol and the H2O are liquids. You can have two outcomes A and B. A) If your limiting reactant is the propan-1-ol, you will have to calculate the excess of moles of O2 and add this to the moles of CO2 produced to calculate then the final volume of gas. In this case the gas will be a mixture of CO2 and the an-reacted O2. B) If your limiting reactant is the oxygen, the only gas present after the reaction will be CO2, the propan-1-ol and the H2O are liquids and will not contribute to the volume of gas. 5) Finally you convert the total moles of gas to volume (V=nRT/P) if you need more detail let me know

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