Reaction taking place is: C 2 H 6 + 3.5O 2 ---> 2CO 2 + 3H 2 O Moles of O 2 pres

Question

Reaction taking place is:

C2H6 + 3.5O2 ---> 2CO2 + 3H2O

Moles of O2 present = Mass/MW = 1/32 = 0.03125

Moles of ethane present = Mass/MW = 2/30 = 0.067

1 mole ethane needs 3.5 moles of oxgen for complete reaction

So,

0.067 moles of ethane will need: 0.067*3.5 = 0.2345 moles oxygen

So,

oxygen is the limiting reagent.

Moles of ethane that react = (1/3.5)*0.03125 = 0.0089 moles

Moles of ethane left behind = 0.067-0.0089 = 0.0581

Moles of CO2 produced = 2*0.0089 = 0.0178

Moles of H2O produced = 3*0.0089 = 0.0267

So,

Moles of N2 present = Mass/MW = 2/28 = 0.0714

Mole fraction of N2 after reaction = 0.0714/(0.0714+0.0267+0.0178+0.0581) = 0.41

Hope this helps !

Explanation / Answer

2.00 g C2H6(g) are ignited in a closed vessel that also contains 1.00 g O2(g) and 2.00 g N2(g). The reaction products are CO2(g) and H2O(g). What is the mole fraction of N2(g) after the reaction has gone to completion?

The answer is 0.41, but I just need to see the steps.

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