Chemistry help!!!

Question

Consider the following reaction taking place in a container fitted with a movable piston.

2 SO₂(g) + O₂(g) → 2 SO₃(g)

Suppose we place two moles each of SO₂ and O₂ in the reaction vessel at 25^oC and adjust the volume to give a total pressure of 1.0 atm. The reaction is ignited by a spark and goes to completion. The temperature is returned to 25^oC. Which of the following BEST describes this system after the reaction is complete?

a O2 is limiting. The volume of the system remains the same.

b SO2 is limiting. The volume of the system remains the same.

c SO2 is limitnig. The volume of the system decreases.

d Neither reactant is limiting. The volume of the system decreases.

Answer 1

Hi Rachel,

The answer is C

First you must determine what the limiting reactant is for this reaction. Since they give you the moles of both reagents, this should be simple. See which mole fraction will produce the least amount of product (SO3). Remember that mole fraction is always = Moles x (Product coefficient/Reactant coefficient).

For O₂: 2moles x (2/1) = 4moles SO3

For SO2: 2moles x (2/2) = 2moles SO3

Therefore, SO₂ is the limiting reactant. So now it's between (B) and (C). If we assume that all the gases are "ideal gases", we can solve this quite easily by comparing the total moles of gas before and after the reaction. Before the reaction there were 4 moles of gas total (2moles SO2 + 2moles O2). After the reaction, all SO2 is depleted (0moles), 2moles of SO3 were created, and 1mole of O2 was used, resulting in 1mole of O2 left in excess. This means there are only 3moles of gases after the reaction (2moles SO3 + 1mole O2). Therefore there was a decrease in volume (C).

I often find my students struggle with knowing how they determine how much excess reagent was used in limiting reactant problems. For determining that 1mole of O2 was used, use the same method we did for determining the limiting reagent, except have the mole fraction be (Coefficient of O2 / coefficient of SO2).

2moles SO2 x (1/2) = 1mole of O2 used in this reaction.

Therefore, of the 2moles of O2 we started with, there was only 1mole of O2 left.

Hope this helps!