Which is correct?

In an experiment, 12.0 dm3 of oxygen, measured under room conditions, is used to burn
completely 0.10 mol of propan-1-ol.
What is the final volume of gas, measured under room conditions?
A 7.20 dm3 B 8.40 dm3 C 16.8 dm3 D 18.00 dm3

Reply 1

charco

Study Forum Helper

Original post by TTY13

... what exactly is the problem?

Have you written out the equation for the reaction?

Reply 2

TTY13

Yes I have. And I got my final answer as 7.20 but the correct answer is 8.40dm^3

Original post by charco

... what exactly is the problem?

Have you written out the equation for the reaction?

Reply 3

charco

Study Forum Helper

Original post by TTY13

Yes I have. And I got my final answer as 7.20 but the correct answer is 8.40dm^3

show your equation

Reply 4

eznot

The answer will depend on what volume you assume one mole of gas takes up, if you assume 24dm3 the 7.2dm3 is the correct answer, if you assume 22.4dm3 you will get an answer of 8.0dm3

Reply 5

TTY13

2c3h7oh+ 9o2 = 6co2 + 8h2o

Original post by charco

show your equation

Reply 6

charco

Study Forum Helper

Original post by TTY13

2c3h7oh+ 9o2 = 6co2 + 8h2o

OK, so from the equation 0.1 mol propan-1-ol makes 0.3 mol of carbon dioxide and to do so burns 0.45 mol of oxygen
as 1 mol = 24 dm³ @RTP
This means that carbon dioxide = 7.2 dm³ is created and 10.8 dm³ of oxygen is used.
Initially there were 12 dm³ of oxygen therefore 1.2 dm³ of oxygen remain.
Add this to the carbon dioxide formed and the total volume is 7.2 + 1.2 = 8.4 dm³

Reply 7

furqan56

the final volume is 8.4 dm^3 because along with 7.3dm^3 of CO2, the unreacted o2, which is 1.2dm^3 when calculated, also is counted.

Reply 8

Goodnews1

Fantastic solutions.

Reply 9

akkz73

Original post by TTY13

1 mol propan-1-ol requires 4.5 mol oxygen for complete combustion. So 0.1 propan-1-ol mol needs (0.14.5)=0.45 mol oxygen. 12 dm³ i.e. 0.5 mol oxygen is used in the reaction, which means (0.5-0.45)=0.05 mol oxygen is left in the end. The amount of CO₂ is 0.3 mol in the end (1 mol propan-1-ol produces 3 mol CO₂, 0.1 will produce (0.13)=0.3 mol CO₂). So, in total, 0.3+0.05=0.35 mol of gas is present in the end, which is 0.3524=8.4 dm³