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Hi, there is a question I'm in doubt with, I was wondering if anyone could help

10cm^3 of a hydrocarbon, CxHy were exploded with an excess of oxygen. A contraction of 25cm^3 occurred. On treating the products with aqueous NaOH, a further contraction in volume of 40 cm^3 occurred. Deduce the values of 'x' and 'y'.

Thanks
Hi @beautyindevil i got the value of x=4 and y=3. My workings are as follows:

Reaction 1 : CxHy + Z O2 --> x CO2 + y/2 H2O

Volume on left = 10 + V (unknown) O2
Volume on right = volume on left -25
therefore: 10 + V --> (10 +V) -25

As we know H2O isn't a gas we can attribute all the volume on the right hand side to CO2. Simplified the volume is (V-15)cm3. We haven't quite got enough information to calculate volume so must consider the next equation.

CO2 + 2NaOH --> Na2CO3 + H2O
As we know the two products on the RHS aren't gases, they occupy 0 volume. This means the 40cm3 contraction is the total volume of the LHS reactants. As NaOH is a solution then must be all CO2, so we have 40cm3 CO2.

We can put this value back in the first equation, as we know CO2 is also the only gas and must also be 40cm3

We can now calculate the moles of xCO2 using the ideal gas equation by dividing by 24,000 cm3
55/24000 = 1.67 x 10-3 moles
We also know the moles of CxHy reacted:
10/24000 = 4.2 x 10-4 moles
We can now calculate the volume of O2 reacted as we know V - 15 = 40 so V=55cm3
55/24000 = 2.3 x 10-3 moles


dividing the number of moles of CO2 by moles of CxHy gives us 4.

C4Hy + ZO2 --> 4CO2 + y/2 H2O

We can also calculate the O2:C4Hy by dividing by moles above:
2.3 x 10-3 / 4.2 x 10-4 = 5.5 moles. We now have

C4Hy + 5.5 O2 --> 4CO2 + y/2 H2O

1.5 O is needed to balance the RHS so y = 3

Hope this helps. Let me know if you have any further questions.

Rob
PhD Chemistry student
Student ambassador
Thank You so much Rob!
I'm so sorry, I didn't see the reply. I will go back and re-try the question again to see if the answers I get match yours.
Thanks

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