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a 125 torr mixture of c2h4(g) and c3hg(g) at a given temperature was burned in excess oxygen in a rigid, sealed container. the product c02( g) was collected under the same initial reaction conditions and determined to be 280 torr. what was the mole fraction of c2h4(g) in the original mixture

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Answer:

0.24

Explanation:

C2H4(g) + 3O2(g) ------> 2CO2(g) + 2H2O(g)

C3 H8 (g) + 5O2 (g) → 3CO2(g)  +4H2 O(g)

Initially,

PV=nRT

125×V=(a+b)RT .....(i)

After combustion, pressure is due to the total moles of CO2  

280×V=(3a+2b)RT .....(ii)

Dividing equation ii by i

280/125 = 3a+2b/a+b

2.24 = 3a+2b/a+b

2.24a + 2.24b =3a+2b

2.24b - 2b = 3a - 2.24a

0.24b = 0.76a

b= 3.2a

mole fraction of C3H8 = a/a + b = a/a + 3.2a = 1/4.2 = 0.24

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