Question

at 1.00 atm and 0 c a 5.04 mixture of methane and propane was burned producing 20.9 g of co2. what was the mole fraction of each gas in the mixture? assume complete combustion

Answer 1

Mol of CO2 produced :
Molar mass CO2 = 44g/mol
17.2g CO2 = 17.2/44 = 0.391 mol CO2 produced .

Mol of gas burned :
At STP 1mol gas = 22.4L
5.04L gas = 5.04 /22.4 = 0.225 mol mixed gases

Let X mol of methane be used
Then (0.225-X) mol propane was burned

1mol CH4 will produce 1 mol CO2
1 mol C3H8 will produce 3 mol CO2

X mol CH4 will produce X mol CO2
(0.225-X) mol C3H8 will produce 3( 0.225-X) mol CO2 = (0.675 - 3X) mol CO2

Mol CO2 from CH4 + mol CO2 from C3H8 = 0.391 mol CO2 in total

X + 0.675 - 3X = 0.391
-2X = -0.675 + 0.391
-2X = -0.284
X = 0.142

You have 0.142 mol CH4 and 0.225-0.142 = 0.083 mol C3H8

Calculate mol fraction:
summarise data:
mol CH4 = 0.142
Mol C3H8 = 0.083
total moles = 0.225

mol fraction CH4 = 0.142/0.225 = 0.631
Mol fraction C3H8 = 0.083/ 0.225 = 0.369

Answer 2

Yes that process worked. I had figured out most of it, but not the second half I could not. It would be nice to not this is different then my question. 17.9g co2 is not 20.9g. So when you answer a quick note of that difference would help.