Question

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)

Answer 1

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:

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