stociometry calculate grams

Question

Ethanol (C₂H₆O) is combusted in air according to the following reaction: C₂H₆O(l) + O₂(g) → CO₂(g) + H₂O(l) How many grams of water would be produced by the complete combustion of 12.10 moles of ethanol in the presence of excess oxygen?

Answer 1

Hello!

Lets start by balencing your chemical equation :)

C₂H₆O(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)

Note that when writing an equation that each reagent is in units of mole.

We know that we have 12.10 moles of C₂H₆O(l).

Because we have excess oxygen, we don't have to worry about finding a limiting reactant. :D

Lets convert 12.10 moles of C₂H₆O(l) into moles of H₂O. This can be done using the balenced chemical equation as it can be seen that for every 1 mole of C₂H₆O(l) that we have, we have 3 moles of H₂O(l). (The 1 and the 3 are the coefficents in the chemical equation)

(12.10mol C₂H₆O(l)) x (3mol H₂O(l)/1 mol C₂H₆O(l)) = 36.3mol H₂O

Now, we know how much water we are making using the ethanol, great! However, this value is in the unit of mole, and the question wants us to answer in grams. We can convert moles to grams by using the molar mass.

Molar Mass H₂O = 18.02g/mol

(36.3mol H₂O) x (18.02g/mol) = 654.13g H₂O

Final Answer: 654.13g H₂O are produced in the complete combustion of 12.10 moles of C₂H₆O(l).

Hope this was helpful!!