Amixture of c2h2 and ch4 has a total mass of 230.9 g. when this mixture reacts completely with excess oxygen, the co2 and h2o products have a combined mass of 972.7 g. what mass of ch4 was present in the initial mixture?

Interesting problem. Thanks for posting.

C2H2 + (3/2)02 >  H2O + 2CO2
CH4 +  2O2 > 2H2O + CO2

The molar mass of C2H2 = 2*12 + 2*1 = 26
The molar mass of CH4 = 1*12 + 4*1 = 16

The number of moles of C2H2 = x
The number of moles of  CH4 = y
26x + 16y = 230.9 grams

For water we get (from the C2H2). Water has a molar mass of 2*1 + 16 = 18

x*18 See the balanced equation to see what it is the same number of moles as C2H2
From the methane we get
y*18
2*y* 18. Again see the balanced equation to see where that 2 came from.
18x + 36y is the total amount of water. 

Now for the CO2. CO2 has a molar mass of 12 + 2*16 = 44
From C2H2  we get 2*44*x = 88x grams of CO2
From CH4 we get 1*y*44 grams of CO2
88x + 44y for CO2

Now we total to get the grand total of water and CO2
18x + 44y + 88x + 44y = 972.7 grams total.
106x + 88y =  972.7

Two equations, two unknowns, we should be able to solve this problem
26x + 16y = 230.9
106x + 88y = 972.7

I'm not going to go through the math unless you request me to do so. 
x = 8.03 moles
y = 1.38 moles 

The initial amount of C2H2 was 8.03 * 26 = 208.78
The initial amount of CH4 was 16*1.38 = 22.08
The total (as a check is 230.86 which is pretty close to the given amount.
So Methane's mass in the initial givens was 22.08 grams.