if 7.o moles of sulfur atoms and 10 moles of oxygen molecules are combined to form the maximum amount of sulfur trioxide, how many moles of which reactant remain unused at the end?
2S + 3O2 -> 2SO3
This example should help you.
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm
2S + 3O2 ==> 2SO3
7.0 mols S x (2 mol SO3/2 mol S) = 7.0 mols SO2 formed if we had 7.0 mols S and all the O2 needed.
10 mols O2 x (2 mol SO3/3 mol O2) = 6.67 mols SO2 formed if we had 10 mols O2 and all the S needed.
In limiting reagent questions the smaller value is ALWAYS the correct one to choose; therefore, the limiting reagent is O2 and some S will be remain unreacted.
To determine the reactant that remains unused at the end, we first need to find the stoichiometric ratio between sulfur atoms and oxygen molecules in the reaction.
The balanced equation for the formation of sulfur trioxide (SO3) is:
2 SO2 + O2 → 2 SO3
From the balanced equation, we can see that for every 2 moles of sulfur dioxide (SO2) consumed, 1 mole of oxygen (O2) is consumed, and it produces 2 moles of sulfur trioxide (SO3).
Given:
- 7.0 moles of sulfur atoms
- 10 moles of oxygen molecules
The molar ratio between sulfur atoms and sulfur dioxide is 2:2, which simplifies to 1:1. So, we have an equal amount of sulfur dioxide and sulfur atoms.
Since there are 7.0 moles of sulfur atoms, there must also be 7.0 moles of sulfur dioxide.
Now, let's calculate the amount of oxygen molecules required for the complete reaction:
- The stoichiometric ratio between oxygen and sulfur dioxide is 1:2, meaning 1 mole of oxygen is required for every 2 moles of sulfur dioxide.
- Since we have 7.0 moles of sulfur dioxide, we need half that amount (7.0/2 = 3.5 moles) of oxygen molecules for complete reaction.
However, we have 10 moles of oxygen molecules, which is more than enough for the reaction. Therefore, the sulfur dioxide will be fully consumed, but there will be unused oxygen molecules remaining.
Answer:
At the end of the reaction, 7.0 moles of sulfur dioxide will be fully consumed, and 10 - 3.5 = 6.5 moles of oxygen molecules will remain unused.
To find out how many moles of each reactant remain unused at the end, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and restricts the amount of product that can be formed.
First, we need to determine the balanced chemical equation for the reaction between sulfur and oxygen to form sulfur trioxide:
2 SO2 + O2 → 2 SO3
From the balanced equation, we can see that 2 moles of SO2 react with 1 mole of O2 to produce 2 moles of SO3.
Given:
- 7.0 moles of sulfur atoms (S)
- 10.0 moles of oxygen molecules (O2)
We need to convert the number of moles of sulfur atoms to moles of SO2 by using the molar ratio from the balanced equation:
7.0 moles S × (1 mole SO2 / 1 mole S) = 7.0 moles SO2
Now we compare the moles of SO2 to the moles of O2 to determine the limiting reactant. Assuming all the sulfur atoms react, we need to convert the moles of O2 to moles of SO2:
10.0 moles O2 × (2 moles SO2 / 1 mole O2) = 20.0 moles SO2
Since we only have 7.0 moles of SO2 available, it is the limiting reactant. This means that all 7.0 moles of SO2 will react completely with 3.5 moles of O2 (according to the stoichiometric ratio of 2:1 between SO2 and O2).
To find the moles of O2 remaining, we subtract the moles of O2 used from the initial moles of O2:
10.0 moles O2 - 3.5 moles O2 = 6.5 moles O2 remain unused.
Therefore, at the end, 6.5 moles of O2 remain unused.