Answer:
88 grams of CO₂ are produced in this combustion.
Explanation:
The reaction for this combustion is:
C₂H₆O (l) + 3O₂(g) → 2CO₂(g) + 3H₂O (l)
Let's calculate the moles
Moles = mass / molar mass
Moles Ethanol = 46 g / 46 g/m = 1 mol
Moles O₂ = 96 g / 32 g/m = 3 moles
Both quantity are the same for both reactant, so the moles produced of products are the same as stoichiometry.
There is also no reactant in excess, either limiting reactant.
We have 2 moles of CO₂ produced.
Mass = Molar mass . Moles
Mass = 44 g/m . 2 m = 88 g