At 1.00 atm and 0 °C, a 5.04 L mixture of methane (CH4) and propane (C3H8) was burned, producing 16.5 g of CO2. What was the mole fraction of each gas in the mixture? Assume complete combustion.

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anonymous · Answer

I cannot seem get the right answer can anyone see where i'm going wrong?

Q.¬2  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 

have 0.142 mol CH4 and 0.225-0.142 = 0.083 mol C3H8 

mol fraction: 
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