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3 years ago

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A 200.0mL closed flask contains 2.000mol of carbon monoxide gas and 2.000mol of oxygen gas at the temperature of 300.0K. How man

y moles of oxygen have to react with carbon monoxide in order to decrease the overall pressure in the flask by 10.00%? Assume ideal gas behavior. The reaction of carbon monoxide and oxygen gas is described by the following equation. 2CO(g)+O2(g)⟶2CO2(g)

1 answer:

emmasim [6.3K]3 years ago

7 0

Answer:

0,400 moles of Oxygen

Explanation:

Using PV = nRT it is possible to obtain the initial pressure of the flask before the reaction thus:

P = nRT/V

Where:

n are moles (4,000 moles, 2,000 of CO and 2,000 moles of H₂O)

R is gas constant (0,082atmL/molK)

T is temperature (300,0K)

V is volume (0,2000L)

Replacing, <em>P = 492,0 atm</em>

If you want to decrease the pressure in 10,00%, the final pressure must be:

492,0atm - 49,2 atm = 442,8 atm

Solving under the same conditions with this pressure, moles must be:

n = PV/RT

<em>n = 3,600 total moles</em>

In the reaction:

2CO(g) + O₂(g) ⟶ 2CO₂(g)

The moles you will have are:

CO: 2,000 moles - 2X

O₂: 2,000 moles - X

CO₂: 2X

<em>Where X are moles that react</em>

Thus, total moles are:

4,000moles - X = 3,600 moles

X = 0,400 moles

That means that moles of oxygen that have to react are <em>0,400 moles of Oxygen</em>

<em />

I hope it helps!

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