Solve Stoichiometry Problem with 2.0 mol O2 & Excess S

[CAF1119]

I need some help with a stoichiometry problem from my basic Chem class.
Here it is: For the reaction: 2S(s) + 3O2 -> 2SO2(g)
How many moles of SO3 will be produced from 2.0 mol O2 and excess S?

This seems like a very simple conversion from moles O2 to moles SO3 - to which I have an answer of 1.3 mol SO3, but I don't think it is correct (I have no answer key). I am getting confused with the "and excess S" part of the question and think that more needs to be done? Any help?

 

[Cesium]

I need some help with a stoichiometry problem from my basic Chem class.
Here it is: For the reaction: 2S(s) + 3O2 -> 2SO2(g)
How many moles of SO3 will be produced from 2.0 mol O2 and excess S?

I'll asume that was just a typo and should be SO3 as the product instead of SO2.

This seems like a very simple conversion from moles O2 to moles SO3 - to which I have an answer of 1.3 mol SO3, but I don't think it is correct (I have no answer key). I am getting confused with the "and excess S" part of the question and think that more needs to be done? Any help?

Your answer is correct. The fact that S is in excess just means that all 2.0 moles of O2 will react with the S producing SO3. You'll end up with an unknown amount of excess S after the reaction has taken place, but you know that all 2.0 moles of O2 have reacted.

Say the problem read like this: How many moles of SO3 will be produced from 2.0 mol O2 and 1.0 mol S?

In this case, you must find which reactant is the limiting reactant and which reactant is in excess. Using the chemical equation (2S(s) + 3O2 -> 2SO3(g)), you can see that for every 2 moles of S, 3 moles of O2 react. Since you have only 1.0 mol S, you know that only 1.5 moles O2 will react. Therefore you will have 0.5 mole O2 left over (2.0-1.5=0.5). O2 is in excess and S is the limiting reactant. Since S is the limiting reactant, all 1.0 mol S reacts to form 1.0 mol SO3.

 

[Blueshift5]

Based on the given information, it appears that the reaction is between 2 moles of sulfur (S) and 3 moles of oxygen (O2), and it produces 2 moles of sulfur dioxide (SO2). However, we are given 2.0 moles of O2 and excess S, meaning that there is more than enough S present to react with the O2.

To solve this stoichiometry problem, we can use the given mole ratio from the balanced equation (2 moles O2 : 2 moles SO2) to calculate the amount of SO2 produced from 2.0 moles of O2. This gives us 2.0 moles SO2.

Since there is excess S present, we can assume that all of the O2 will be used up in the reaction and there will be leftover S. Therefore, the amount of SO2 produced will be limited by the amount of O2, and we can use the mole ratio from the balanced equation (2 moles SO2 : 3 moles O2) to calculate the amount of SO3 produced. This gives us 1.33 moles SO3 as the final answer.

In summary, the correct answer to the stoichiometry problem is 1.33 mol SO3 produced from 2.0 mol O2 and excess S. It is important to pay attention to the given information and use the proper mole ratios from the balanced equation to accurately solve stoichiometry problems.