Question

The combustion of ethanol (C2H6O) with oxygen yields CO2 and H2O. When burned as an alternative, ‘greener’ fuel it is desirable to avoid left over reactants. Suppose you react 0.15 mol ethanol with 0.25 mol oxygen in a closed container. Determine if any reactants be left over and how much (in moles)?

Answer 1

0.07 moles of
`C_(2)H_(6)O` are left over

Step-by-step explanation:

1. Write down the balanced chemical equation for the combustion of ethanol with oxygen:

`C_(2)H_(6)O+3O_(2)=2CO_(2)+3H_(2)O`

2. Find the limiting reagent to know which reagent is left over:

To find the limiting reagent take the number of moles given by the problem and divide this number between the stoichiometric coefficient from the balanced chemical equation. The smallest number will be the limiting reagent.

- For the ethanol :

`(0.15)/(1)=0.15`

- For the oxygen:

`(0.25)/(3)=0.08`

Therefore, the oxygen is the limiting reagent, that is the ethanol is the compound that will be left over.

3. Find the moles of ethanol left over.

- First calculate how many moles of ethanol reacted with the oxygen:

`0.25molesO_(2)*(1molC_(2)H_(6)O)/(3molesO_(2))=0.08molesC_(2)H_(6)O`

So 0.08 moles of
`C_(2)H_(6)O` reacted with the oxygen.

- Then subtract the moles of ethanol that reacted with the oxygen from the total moles of ethanol in the reaction:

`0.15molesC_(2)H_(6)O-0.08molesC_(2)H_(6)O=0.07molesC_(2)H_(6)O` are left over.