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 [Solved]

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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

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

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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

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

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)

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

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=

This post is last updated on hrtanswers.com at Date : 1st of September – 2022