A steel cylinder contains 5.00 mol of graphite (pure carbon) and 5.00 mol of O2. The mixture is ignited and all the graphite reacts. Combustion produced a mixture of CO gas and CO2 gas. After the cylinder has cooled to its original temperature, it is found that the pressure in the cylinder has increased by 16.4%. Calculate the mole fractions of CO, CO2, and O2 in the final

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jufsanabriasa jufsanabriasa · Answer 1 · 2020-09-11T03:30:41+02:00

Answer:

O₂ = 0.141

CO = 0.282

CO₂ = 0.577

Explanation:

The reaction of graphite with Oxygen is:

C(s) + O₂(g) → CO₂(g) + CO

All 5 moles of C(s) reacts producing CO₂ and CO. You can write:

5 moles = CO₂ + CO (1)

And, as pressure increased by 16.4% and initial moles of oxygen are 5.00 moles. Final moles (Moles of three gases) are:

5.00moles + 5.00moles*16.4% =

5.82 moles = O₂ + CO₂ + CO (2)

Also, there are 10 moles of oxygen in the container, that is:

10 moles = O₂*2 + CO₂*2 + CO (3)

From (2) and (1), moles of O₂ are:

O₂ = 0.82 moles

Replacing in (3)

10 mol = 0.82*2 + CO₂/2 + CO

8.36 moles = CO₂*2 + CO

5 moles = CO₂ + CO

By the sum of these equations:

3.36 moles = CO₂

And

5.0 moles - 3.36 moles = CO

1.64 moles = CO

As total moles are 5.84 moles, mole fraction of the gases is: