3
$\begingroup$

$60\ \mathrm{cm^3}$ of oxygen was added to $10\ \mathrm{cm^3}$ of a gaseous unsaturated hydrocarbon. After explosion and cooling to room temperature, the residual gases occupied $45\ \mathrm{cm^3}$. What volume of carbon dioxide was formed in the reaction?

I’ve tried to make equation out of it:

$$\ce{(CH2)$_n$ + O2 -> CO2 + H2O}$$

My problem now is I don’t know how to proceed to the next step. How do I even find the moles? Is the equation correct? How to stabilize the equation? Please, guide me step by step. It’s fine if you don’t want to give me the final or exact answer. I just need someone to guide me how to solve this problem. Thanks in advance.

$\endgroup$
  • $\begingroup$ You are missing 2 hydrogens; also, indicate which state each of the compounds are in. $\endgroup$ – LDC3 Jul 18 '15 at 13:19
4
$\begingroup$

First step: balance your equation.

$\ce{(CH2)$_n$ + O2 -> CO2 + H2O}$ is not balanced, this is:

$$\ce{(CH2)_{\it n} + \frac{3 }{2}{\it n}~O2 -> {\it n}~ CO2 + {\it n}~ H2O}$$

Second step: perform mole balances on relevant species.

The balanced equation has $1 + \frac{3}{2}n$ moles of gas reacting, and $2 n$ moles of gas formed. However, the question states that the products are cooled to room temperature, which would eliminate the water from the gas phase. Thus the reaction were are really interested in is

$$\ce{(CH2)_{\it n}_{\rm(g)} + \frac{3 }{2}{\it n}~O2_{(g)} -> {\it n}~ CO2_{(g)} + {\it n}~ H2O}_{(l)}$$

This means that $1 + \frac{3}{2}n$ moles of gas react, and $n$ moles form for each mole of reaction $\xi$. Thus, the reaction causes a loss in moles of gas of is $-(1 + \frac{1}{2}n)$ moles per mol $\xi$ of reaction. If we assume the ideal gas law applies, then we could also say that the loss in volume from the reaction is $(1 + \frac{1}{2}n)~{\mathrm{{cm}^3}}$ per $\xi~{\mathrm{{cm}^3}}$ of reaction.

Thus a total gas balance is:

$$\rm GAS~IN - GAS~OUT = GAS~REACTED $$ $$(60~\mathrm{{cm}^3} + 10~\mathrm{{cm}^3}) - (45~\mathrm{{cm}^3})=(1 + \frac{1}{2}n)~\xi$$

A $\ce{CO2}$ balance is

$$\rm \ce{CO2}~IN - \ce{CO2}~OUT = \ce{CO2}~REACTED $$ $$ 0 - x = - n \xi $$

Third step: assess the resulting equations

These equations are not solvable for $x$ because they depend on both $\xi$ and $n$. That is, there are two equations, and three unknowns. Writing additional balances on oxygen or on fuel will not help because we will not know the gas remaining for either species, so each new equation would introduce new variables. Thus, we will need additional assumptions to solve the problem.

Fourth step: determine additional assumptions.

Probably we should assume that either (a) fuel was in excess or (b) oxygen was in excess, which is equivalent to assuming that (a) oxygen was completely consumed or (b) fuel was completely consumed. At least one of these assumptions makes sense because combustion reactions are essentially irreversible. In my answer I'll assume oxygen was in excess, meaning that the fuel was completely consumed. You should make the alternate assumption and repeat the procedure.

Fifth step: apply additional assumptions.

Assuming fuel was completely consumed, a fuel balance is

$$\rm FUEL~IN - FUEL~OUT = FUEL~REACTED $$ $$\rm (10~\mathrm{{cm}^3}) - 0=\xi $$

This conveniently gives us a solution for $\xi$, the extent of reaction. It is $10~\mathrm{{cm}^3}$.

Sixth step: use new information to solve existing equations

We can substitute that solution for $\xi$ into the total gas balance to get that $1 + \frac{1}{2}n = 3$, which implies that $n=4.$ That and our solution for $\xi$ allows us to solve the problem: if oxygen is in excess, then 40 $\mathrm{{cm}^3}$ of $\ce{CO2}$ were formed.

$\endgroup$
  • 1
    $\begingroup$ Also note that your problem statement just says "unsaturated" hydrocarbon, but it doesn't say how unsaturated the compound is. Your formula for an unsaturated hydrocarbon, $\mathrm{{CH}_2}$, assumes that there is only a single degree of unsaturation. $\endgroup$ – Curt F. Jul 18 '15 at 14:29
  • 1
    $\begingroup$ How can you say GAS IN - GAS OUT = GAS REACTED. If I had the stoichiometric amount of fuel and oxygen, then the GAS REACTED would equal the GAS IN, and by your statement, no $\ce {CO2}$ would be produced. $\endgroup$ – LDC3 Jul 18 '15 at 14:35
  • 1
    $\begingroup$ Good point LDC3. I may have used misleading names. By GAS REACTED, I mean the net change in gas from the reaction. So if there was a stoichiometric mix of fuel and oxygen, there would still be a loss of $(1 + \frac{1}{2}n) \xi$ moles of gas by the reaction, as reflected in the next equation I wrote. $\endgroup$ – Curt F. Jul 18 '15 at 14:39
0
$\begingroup$

First balancing:

2(CH2)+3O2 -> 2(CO2)+2(H2O) 2v +3v -> 2v+2v 3v->60 2v->40

Therefore, the answer should be 40cm^3

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.