A mixture of substance with the formula C
3H
6O and C
3H
8O is burned in the presence of a stoichiometric amount of air (20% oxygen and 80% of nitrogen by volume) in an eudiometer to produce a mixture of gasses with a total volume of 5.432dm
3 at STP. The mixture is then bubbled through Ba(OH)
2 solution, its volume decreases by 15.46%. Calculate the molar ratio of the substances in the beginning mixture.
Attempt:xC
3H
6O+4.5xO
2 3xCO
2+3xH
2O
yC
3H
8O+5yO
2 3yCO
2+4yH
2O
I mark the number of moles of C
3H
6O with
x and of C
3H
8O with
y.
The volume decrease is because of CO
2 whose number of moles is n=5.432*0.1546/22.4=0.0375mole, this means that:
x+y=0.0375The remaining mixture should contain nitrogen (V=4.592dm
3) so the volume of oxygen that was spent was 4 times smaller and its number of moles is: n=4.592/(4*22.4)=0.0512mole, from the reaction equation:
4.5x+5y=0.0512. When this is solved with the first equation (
x+y=0.0375) I get a negative answer.
Then I thought that maybe water was in gas state, too. The first equation is the same, now to the second one. The number of moles of the mixture is equal to the sum of the moles of water and 4*number of moles of oxygen (because there was 4 times more nitrogen in air than oxygen): 4.592/22.4=3x+4y+4*(4.5x+5y), i.e.
21x+24y=0.205. When solved with equation
x+y=0.0375 again a negative answer is obtained, so I need help with this.