Answer to Question #89368 in General Chemistry for Samuel Kwaji

Question #89368
How many dm3 of O2 (312K, 118kPa) is necessary to burn 86.6g of ethanol if we use 2.47% excess R=8.314 J/(mole.K)
1
Expert's answer
2019-05-09T02:21:42-0400

Reaction of ethanol combustion is:

C2H5OH + 3O2 → 2 CO2 + 3H2O


Number of moles of ethanol:

"n(EtOH)= \\frac {m(EtOH)}{M(EtOH)} = \\frac {86.6g}{46\\frac{g}{mole}} = 1.8826 mole"


According to reaction stoichiometry number of moles of oxygen is three times more than of ethanol:

"n(O_2) = 3n(EtOH) = 3\\cdot 1.8826 mole = 5.6478 mole"


Taking into account that oxygen was taken in excess, calculating overall quantity of oxygen:

"n(overall O_2) = 1.0247 n(O_2) = 1.0247\\cdot 5.6478 mole = 5.7873 mole"


Using ideal gas law:

pV=nRT

we can find volume of oxygen in given conditions:

"V(O_2) = \\frac{nRT}{p} = \\frac{5.7873mole \\cdot 8.314 \\frac{J}{mole \\cdot K} \\cdot 312K}{118000Pa} = 0.127 m^3 = 127 dm^3"


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