the combustion of ethanol is as follows;
C₂H₅OH + 3O₂ ---> 2CO₂ + 3H₂O
stoichiometry of C₂H₅OH to CO₂ is 1:2
the mass of ethanol that was used - 46.1 g
Number of ethanol moles used up - 46.1 g/ 46.1 g/mol = 1 mol
according to stoichiometry, when 1 mol of ethanol is used up, 2 mol of CO₂ is produced
Therefore mass of CO₂ produced - 2 mol x 44 g/mol = 88 g
