# At 1.00 atm and 0 °C, a 5.04 L mixture of methane (CH4) and propane (C3H8) was burned, producing 17.2 g of CO2?

At 1.00 atm and 0 °C, a 5.04 L mixture of methane (CH4) and propane (C3H8) was burned, producing 17.2 g of CO2. What was the mole fraction of each gas in the mixture? Assume complete combustion.

X methane=?

X propane=?

Please explain how to do it! I've gotten this question wrong three times already.

• 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

17.2 g of CO2 correspond to 17/44=0.39 moles of CO2

I look at what amount of moles correspond 5.04 L n=pV/RT=1*5.04/(0.082*273.2) =0.225moles

So the mixture correspond to 0.225 moles . if the fraction of CH4 is x , that of C3H8 is 1-x

so moles of CH4 = 0.225*(1-x) and of C3H8 0.225*(1-x)

when burning 1 mole of CH4 gives 1 mole of CO2 and 1 mole of C3H8 3moles of CO2

so 0.39= 0.225x + 0.225*3(1-x)=0.225x+0.675-0.675x=0.675-0.45x

0.45x =0.675-0.39=