A rigid 1.7 L sealed vessel contains 3.3 mol of O2(g), 1.6 mol of CH4(g), and 2.

Question

A rigid 1.7 L sealed vessel contains 3.3 mol of O2(g), 1.6 mol of CH4(g), and 2.3 mol of He(g) has an internal temperature of 78 C.

I can answer a, b, and c which is (a) partial pressure of each gas, (b) total pressure in the vessel, (c) mol fraction of each gas.

But I am stuck on (d) which says "A lab technician ignites the mixture in the flask and the following reaction occurs: CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g). Find the mole fraction of each gas in the mixture after the reaction. (Hint: Use an ICE chart. There is a limiting reactant.)

Explanation / Answer

Clearly,here the reaction stociometric balence shows-

1mole of CH4(g) reacts with 2 moles O2(g) to produce 1 mole of CO2(g) and 2 moles of H2O(g).

Thus clearly by unitary method 1.6moles of CH4(g) reacts with 3.2 moles O2(g) to produce 1.6 mole of CO2(g) and 3.2 moles of H2O(g).

Thus O2 reamins unreacted = (3.3-3.2)=0.1moles

He(g) is inert thus it does not participates in the reaction.

THUs after the completion of the reaction in the vessel there is 1.6 moles of CO2(g), 3.2moles of H2O(g), 0.1moles of unreacted O2(g) and 2.3 moles of He(g).

So, the total no moles after the reaction is = (1.6+3.2+0.1+2.3)=7.2 moles

So the mole fraction of CO2(g) = (1.6/7.2) = 0.222

the mole fraction of H2O(g) = (3.2/7.2) = 0.444

the mole fraction O2(g) = (0.1/7.2) = 0.014

the mole fraction of He(g) = (2.3/7.2) = 0.319

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