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Amount of Substance question

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 O2(g) → 8 CO2(g) + 9 H2O(l)

Calculate the amount, in moles, of gas in the flask after the reaction.

I have sat here trying to do this for 20 minutes and don't get it at all. can someone please explain

This is what the mark scheme says:
M1 amount of CO2 formed in flask = 0.008 mol
Allow ECF from M1 to M2

M2 amount of gas in flask
= 0.0075 (O2) + 0.0080 (M1) = 0.0155 mol

Reply 1

You have 20x the amount of O2 (at the start) compared to how many mol of C8H18 you start with.

1 mol of C8H18 uses up 12 mol of O2 - so 8x the amount of O2 compared to the amount of C8H18 must be left over, unreacted (20 - 12 = 8)

1 mol of C8H18 makes a total of 17 mol of gaseous products (8 + 9 = 17)

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