[Blueberries116]

The problem is as follows:

A mixture of [itex]CH_{4}[/itex] and [itex]C_{2}H_{4}[/itex] is let to pass over red hot [itex]CuO[/itex]. Then [itex]0.6\,g[/itex] of [itex]H_{2}O[/itex] is collected along [itex]1.185\,g[/itex] of [/itex]CO_{2}[/itex]. What will be the composition of the mixture if it is known that the combustion was complete?.

The alternatives given in my book are as follows:

[itex]\begin{array}{ll}
1.&CH_4=51.8\%\,and\,C_2H_4=48.2\,\%\\
2.&CH_4=50\%\,and\,C_2H_4=50\,\%\\
3.&CH_4=33.3\%\,and\,C_2H_4=66.6\,\%\\
4.&CH_4=38.8\%\,and\,C_2H_4=64.2\,\%\\
5.&CH_4=36.3\%\,and\,C_2H_4=63.7\,\%\\
\end{array}[/itex]

I'm not sure exactly how to proceed with this question. Should it be treated as if it is an analysis by combustion?. Or just as a normal problem of stoichiometry?.  Can someone help me to make the equations needed to solve this?. Help please.

[Borek]

Not a best question if you ask me, way too sensitive to the numerical inaccuracies.

But in general it is trivial: write mass (or number of moles) balances for carbon and hydrogen, expressing their amounts in terms of numbers of moles of both gases in the mixture.

[Blueberries116]

Not a best question if you ask me, way too sensitive to the numerical inaccuracies.

But in general it is trivial: write mass (or number of moles) balances for carbon and hydrogen, expressing their amounts in terms of numbers of moles of both gases in the mixture.

Well... I agree with you regarding the trivial thing, but I'm still not sure if what comes below is what you were intending to say.

What I did to solve this was to treat each gas in the mixture as reacting independently for the oxigen from interacting with the red hot [itex]CuO[/itex] as follows:

[itex]CH_{4}+2O_{2} \rightarrow CO_{2}+2H_{2}O[/itex]

With: [itex]x=CH_{4}[/itex]

[itex]x\,g\times\frac{1\,mol\,CH_{4}}{16\,g\,CH_{4}}\times \frac{2\,\,mol\,H_{2}O}{1\,\,mol\,CH_{4}}\times\frac{18\,g}{1\,mol\,H_{2}O}=\frac{18}{8}x\,g\,H_{2}O[/itex]

[itex]x\,g\times\frac{1\,mol\,CH_{4}}{16\,g\,CH_{4}}\times \frac{1\,\,mol\,CO_{2}}{1\,\,mol\,CH_{4}}\times\frac{44\,g}{1\,mol\,CO_{2}}=\frac{44}{16}x\,g\,CO_{2}[/itex]

For:

[itex]C_{2}H_{4}+3O_{2} \rightarrow 2CO_{2}+2H_{2}O[/itex]

With [itex]y=C_{2}H_{4}[/itex]

[itex]y\,g\times\frac{1\,mol\,C_{2}H_{4}}{28\,g\,C_{2}H_{4}}\times \frac{2\,\,mol\,H_{2}O}{1\,\,mol\,C_{2}H_{4}}\times\frac{18\,g}{1\,mol\,H_{2}O}=\frac{36}{28}y\,g\,H_{2}O[/itex]

[itex]y\,g\times\frac{1\,mol\,C_{2}H_{4}}{28\,g\,C_{2}H_{4}}\times \frac{2\,\,mol\,CO_{2}}{1\,\,mol\,C_{2}H_{4}}\times\frac{44\,g}{1\,mol\,CO_{2}}=\frac{88}{28}x\,g\,CO_{2}[/itex]

This is reduced to a system of equations as follows:

[itex]\frac{44}{16}x+\frac{44}{14}y=1.185\,g\,CO_{2}[/itex]

[itex]\frac{18}{8}x+\frac{18}{14}y=0.6\,g\,H_{2}O[/itex]

Solving this system yields:

[itex]x=0.1024\,g\,CH_{4}[/itex]

[itex]y=0.2874\,g\,C_{2}H_{4}[/itex]

Then to calculate the composition of the mixture I would use the molar fraction:

[itex]n_{CH_{4}}=\frac{0.1024}{16}=0.0064\,mol[/itex]

[itex]n_{C_{2}H_{4}}=\frac{0.2874}{28}=0.01026\,mol[/itex]

Then the percentage for composition for [itex]CH_{4}[/itex] would be:

[itex]\frac{n_{CH_{4}}}{n_{CH_{4}}+n_{C_{2}H_{4}}}=\frac{0.0064}{0.0064+0.01026}=0.3841[/itex]

For [itex]C_{2}H_{4}[/itex]:

[itex]\frac{C_{2}H_{4}}{n_{CH_{4}}+n_{C_{2}H_{4}}}=\frac{0.01026}{0.0064+0.01026}=0.6159[/itex]

To which correspond each one as roughly [itex]38.41\%[/itex] and [itex]61.59\%[/itex] respectively for [itex]CH_{4}[/itex] and [itex]C_{2}H_{4}[/itex]. But none of this seem to check with any of the answers. Could it be that my method was wrong or did I overlooked something?.

Perhaps did you intend to say this?. I'm still stuck on this one.

[AWK]

First of all - type of percentage is missing: v/v or m/m?

Set variables as moles of gases - coefficients for equations are given in balanced reactions (only less rounding).
You will get two equations of the type:
ax + by = mass/MW where a and b are coefficients of balanced equations, for water and carbon dioxide, respectively.
The solution needs one subtraction of equations and substitution of the found variable into one of the equations.
From moles, you can easily find v/v percentage. After conversion moles to masses, you can find mass percentages.
Usee mass of water as 0.6000 g for consistency in significant digits during the solution of system of equations.

[Borek]

To which correspond each one as roughly [itex]38.41\%[/itex] and [itex]61.59\%[/itex] respectively for [itex]CH_{4}[/itex] and [itex]C_{2}H_{4}[/itex]. But none of this seem to check with any of the answers. Could it be that my method was wrong or did I overlooked something?

Many ways to skin that cat (some of them can be slightly faster), but your method is perfectly OK.

I got 38.31% for the final result, probably because I used exact molar masses of gases but for speed copied just four digits in calculations. Unfortunately the way the problem is built makes such differences inevitable, best we can do is to choose the answer that is closest to the answer we got.

As AWK stated, it is not clear whether the question asks for v/v or m/m composition, but m/m doesn't produce numbers that are close to any of the answers to choose from.

[mjc123]

To which correspond each one as roughly 38.41% and 61.59% respectively for CH4 and C2H4. But none of this seem to check with any of the answers

It's pretty close to answer 4 if you assume they meant 38.8 and 61.2 %. As written it must be wrong, because it adds up to more than 100%.

[Blueberries116]

To which correspond each one as roughly [itex]38.41\%[/itex] and [itex]61.59\%[/itex] respectively for [itex]CH_{4}[/itex] and [itex]C_{2}H_{4}[/itex]. But none of this seem to check with any of the answers. Could it be that my method was wrong or did I overlooked something?

Many ways to skin that cat (some of them can be slightly faster), but your method is perfectly OK.

I got 38.31% for the final result, probably because I used exact molar masses of gases but for speed copied just four digits in calculations. Unfortunately the way the problem is built makes such differences inevitable, best we can do is to choose the answer that is closest to the answer we got.

As AWK stated, it is not clear whether the question asks for v/v or m/m composition, but m/m doesn't produce numbers that are close to any of the answers to choose from.

I thought that when no any other specification is given then what is assumed is $w/w$ or percentage by mass. But on any case, the percentage which I obtained is due molar fractions, not the weight obtained from solving the system of equations. As it was mentioned above this post by @mjc123 the second digit after $6$ which is $1$ resembles very closely to $4$ so I believe there was a misprint or some error in printing. But so far there is some consensus that the answer it must be the fourth option. One aspect which was bothering me was why I couldn't obtain an answer which matched the two given numbers so I thought that it was due my miscalculation. But it looks that the logic which I used was right.

[AWK]

With the chemical calculator (exact atomic masses) I got exactly the same result as Borek with his computer program.

[MNIO]

Let's see if I can add to the confusion

********
you have these reactions
  1 CH4 + 4 CuO ---> 1 CO2 + 2 H2O + 4 Cu(s)
  1 C2H4 + 6 CuO ---> 2 CO2 + 2 H2O +6 Cu(s)
and if we assume we have X mol CH4 and Y mol C2H4
  X CH4 + 4X CuO ----> X CO2 + 2X H2O + 4X Cu(s)
  Y C2H4 + 6Y CuO ---> 2Y CO2 + 2Y H2O + 6Y Cu(s)
adding
  X CH4 + Y C2H4 + (4X+6Y) CuO ---> (X + 2Y) CO2 + 2(X+Y) H2O + (4X+6Y) Cu(s)

you collect
  0.6g H2O
  1.185g CO2

*** solution ***
steps
  (1) setup linear equations relating X & Y to moles CO2 and moles H2O
  (2) solve the system of 2 linear equations form mol CH4 and mol C2H4
  (3) simplify the ratios and generate mole and mass fractions

step (1)
    from the coefficient of CO2 in the combined equation and the mass of CO2 collected
         X + 2Y = 1.185g CO2 * (1 mol CO2 / 44.01g CO2)
         X + 2Y = 0.026926
    the coefficient of CO2 in the combined equation and the mass of H2O collected
        2X + 2Y = 0.6g H2O * (1 mol H2O / 18.02g H2O)
        2X + 2Y = 0.033296

step (2)
    solve this system
         X + 2Y = 0.026926
        2X + 2Y = 0.033296
    1st*(-1) + the 2nd gives
          X + 0 = 0.006371
    solving the first for Y
          Y = (0.026926 - 0.006371)/2 = 0.010278

step (3)
    since we want mole %'s, let's do this
         mole % CH4 = 100% * (0.006371 / (0.006371 + 0.010278) = 38.3%
         mole % C2H4 = 100% - 38.3% = 61.7%
    and if you want mass %'s
         mass % CH4 = 0.383*16.04 / (0.383*16.04 + 0.617*28.05)*100% =26.2%
         mass % C2H4 = 73.8%

*********
now let's look at your choices
  (1) #4 is the only answer close to my calc for % CH4 by mole. 
  (2) #4 has a problem.... 38.8% + 64.2% ≠ 100.0%
  (3) your mass of H2O has but 1 sig fig.   that cannot be correct and all the calcs showed need to
        be redone with the correct sig figs.

[AWK]

(2) #4 has a problem.... 38.8% + 64.2% ≠ 100.0%

It is the evident printing error - it should be 61.2

[MNIO]

probably.  but there is still the issue with 0.6g H2O having only 1 sig fig while the rest of the numbers in the problem statement / solution have 3 or 4 sig figs.

[AWK]

I pointed this out earlier. When we do real analyses, we usually get measurement results with similar accuracy.