Questions thread

[ttn]

Hi, can someone please help me out with this qn:

1. The general formula for alkyne is CnH2n-2. When 2 mole of a particular alkyne are completely oxidised, 11 mole of oxygen gas is  required. What is the number of carbon atoms present in this alkyne?

Another similar one:

2. 6.5 L of oxygen is required to completely convert 1.0 L of a gaseous hydrocarbon fuel to carbon dioxide and water. If the volumes were measured at the same temperature and pressure. What would be the hydrocarbon?

Thanks.

[ttn]

Please, anyone?

[Mao]

1. Complete oxidation = combustion. In combustion, C --> CO2 and H --> H2O. Therefore, the number of moles of O2 required per mole of alkyne would be n(O2) = n(C)+0.25n(H) = n(C) + 0.25(2n(C)-2) = 1.5n(C) -0.5

Solving 1.5n(C)-0.5=5.5 gives n(C) = 4. The alkyne is C4H6 (balance the combustion equation and verify)

2. We can convert volume ratio to mole ratio. That is, 6.5 mol of O2 is required to combust 1.0 mol of hydrocarbon. By guesswork, this is C4H10.

If you want a systematic method, n(O2) = 1.5n(C)-0.5 for alkynes, 1.5n(C) for alkenes and 1.5n(C)+0.5 for alkanes. Solving these three expressions for n(O2)=6.5 will only give one integer answer, which is the correct answer.

[ttn]

Thanks a lot for the help, but in question 1, I don't really understand how you got:

n(O2) = n(C)+0.25n(H)

[Mao]

CxHy + zO2 --> xCO2 + y/2 H2O

Therefore z = x + y/4

[ttn]

Ohhh, thanks a lot.