Caitlin asked in Science & MathematicsChemistry · 1 month ago

# Stoichiometry with gas homework help?

2CO(g) + O2(g) --> 2CO2(g)

A 200 ml closed flask contains 2 mol of carbon monoxide gas and 2 mol of oxygen gas at the temperature of 300k. How many moles of oxygen have to react with carbon monoxide in order to decrease the overall pressure in the flask by 10 %? Assume ideal gas behavior.

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• 1 month ago

Supposing the temperature remains constant, by supposing the heat which is generated by the reaction is transmitted to the surroundings, which is to say the flask is not insulated.

Since the volume and temperature remain constant, the total number of moles is proportional to the pressure.  If the pressure is decreased by 10%, then the total number of moles is also decreased by 10%.

Since we start with a total of 4 moles of gases, the final state will be when there is a total of 4 - 10% = 3.6 moles of gases.

2 CO(g) + O2(g) → 2 CO2(g)

Before the reaction begins there is an excess of 1 mol of oxygen, so that 1 mol of oxygen will also be present, unchanged, throughout the reaction.  So of the 3.6 target moles of gases, 1.0 mole will always be unreacted oxygen, and the number of moles of gases involved in the reaction will be 3.0 moles before the reaction begins and 2.6 moles at the desired end.

Looking at the balanced equation, for every 1 mole of O2 that reacts the total number of moles goes from 3 moles to 2 moles.  So for the reaction at hand to stop at 2.6 total moles the reaction must be only 40% complete, and only 40% of the O2 which is not part of the excess will react.  40% of the 1 mol of oxygen which could react is 0.4 moles of O2, which is the answer to this question.

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Checking:

(0.4 mol O2 reacted) x (2 mol CO2 / 1 mol O2) = 0.8 mol CO2 produced

(0.4 mol O2 reacted) x (2 mol CO / 1 mol O2) = 0.8 mol CO consumed

(2 mol CO initially) - (0.8 mol CO consumed) = 1.2 mol CO remaining

(2 mol O2) - (0.4 mol O2 consumed) = 1.6 mol O2 remaining

0.8 mol + 1.2 mol + 1.6 mol = 3.6 mol total final

(4.0 mol total initially) - (3.6 mol total final) = 0.4 mol consumed

(0.4 mol consumed) / (4.0 mol total initially) = 0.10 =

10% decrease in moles, which means a 10% decrease in pressure