AQA Chemistry exam question from AS paper 2 2021

06.2. In the second experiment, another flask is used for a combustion reaction.
Method
• Remove all the air from the flask.
• Add 0.0010 mol of 2,2,4-trimethylpentane (C8H18) to the flask.
• Add 0.0200 mol of oxygen to the flask.
• Spark the mixture to ensure complete combustion.
• Cool the mixture to the original temperature.
The equation is
C8H18(g) + 12 1
2
O2(g) ⟶ 8 CO2(g) + 9 H2O(l)
Calculate the amount, in moles, of gas in the flask after the reaction.

- I understand how to get 0.0080 moles of CO2 but I am unsure how to get the answer of 0.00155, could someone please explain the reason behind this?

Reply 1

charco

Study Forum Helper

Original post by Slugzie(:

06.2. In the second experiment, another flask is used for a combustion reaction.
Method
• Remove all the air from the flask.
• Add 0.0010 mol of 2,2,4-trimethylpentane (C8H18) to the flask.
• Add 0.0200 mol of oxygen to the flask.
• Spark the mixture to ensure complete combustion.
• Cool the mixture to the original temperature.
The equation is
C8H18(g) + 12 1
2
O2(g) ⟶ 8 CO2(g) + 9 H2O(l)
Calculate the amount, in moles, of gas in the flask after the reaction.
- I understand how to get 0.0080 moles of CO2 but I am unsure how to get the answer of 0.00155, could someone please explain the reason behind this?

You have the balanced equation:
C₈H₁₈(g) + 12.5O₂(g) ⟶ 8 CO₂(g) + 9 H₂O(l)
You are told that there is complete combustion, i.e. the oxygen is in excess.
From the stoichiometry, 0.001 mol of 2,2,4-trimethylpentane will produce 0.008 mol of carbon dioxide (water is a liquid and can be ignored)
The oxygen needed is 0.001 x 12.5 = 0.0125
The available oxygen is 0.02, therefore there is an excess of 0.02 - 0.0125 = 0.0075 mol
Therefore the total mol present after reaction = 0.008 + 0.0075 = 0.0155 mol