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.
