Gchem PV=nRT

[keliao]

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am getting a headache trying to solve this problem.

Benzene, C6H6, can be burned in oxygen according to the equation: 2C6H6 (l) + 15O2 (g) --> 12CO2 (g) + 6H2O(g). If 5.0 L of oxygen measured at 1 atm and 273K were required to burn a given amount of benzene, the 1atm and 273K volume of CO2 formed would be ? answer is 4L.

 

[tranv117]

am getting a headache trying to solve this problem.

Benzene, C6H6, can be burned in oxygen according to the equation: 2C6H6 (l) + 15O2 (g) --> 12CO2 (g) + 6H2O(g). If 5.0 L of oxygen measured at 1 atm and 273K were required to burn a given amount of benzene, the 1atm and 273K volume of CO2 formed would be ? answer is 4L.

The question is saying that the reaction occurs at STP. meaning 1 mole of O2 gas will contain the same volume as 1 mole of CO2. This then becomes a matter of ratios. At constant T and P, Avogadros Law comes into play. n1/V1 = n2/V2. 15 mole O2/5L = 12 mole CO2/X. solve for X to get the volume of CO2.

 

[keliao]

thank you so much.