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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