In a combustion reaction, 46.0 of ethanol reacts with 96.0 of oxygen to produce water and carbon dioxide. If 54.0g of water is produced, how much carbon dioxide is produced?

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