Question

A mixture of C3H8 and C2H2 has a mass of 2.6g . It is burned in excess O2 to form a mixture of water and carbon dioxide that contains 1.6 times as many moles of CO2 as of water.


Find the mass of C2H2 in the original mixture.

Answer 1

Two equations in two unknowns. Solve them simultaneously. I think the chemistry is straight forward but the math may be a little tedious.
The combustion equations are as follows:
C3H8 + 5O2 ==> 3CO2 + 4H2O
2C2H2 + 5O2 ==> 4CO2 + 2H2O
Confirm that those are correct.

Let x = mass C3H8
and y = mass C2H2
---------------------
x + y = 2.6
Here is how you get the second equation.
mols CO2 = 1.6*mols H2O
mols CO2 = [(3x/44) + (4x/44)]
mols H2O = [(4y/2*26) + (2y/2*26)]
note: 44 is molar mass C3H8
and 26 is molar mass C2H2. All I've done is convert x grams CO2 and H2O to mols CO2 and H2O.
So the second equation is
[(3x/44) + (4x/44)= = 1.6*[(4y/2*26) + (2y/2*26)] and that can be simplified some to
(7x/44) =1.6*(6y/52).
Find y, then
%C2H2 = (y/2.6)*100 = ?
You should confirm all of this.

Answer 2

This solution is not right. I tried it several times and got the wrong answer
I don't need the percent of c2h2, i want the final mass.