A researcher found that 11.56 g of lead sulfide, PbS, were formed when each of the following occurred:
10.0 g of lead reacted with 1.56 g of sulfur
10.0 g of lead reacted with 3.00 g of sulfur
18.0 g of lead reacted with 1.56 g of sulfur
Explain these observations in terms of the concept of limiting reactant.
Convert 10.0 g Pb to mols.
Convert 1.56 g S to mols.
Convert mols Pb and mols S to mols PbS, then convert to grams PbS for Pb AND for S. You will find that, within reasonable error, each will produce approximately the same amount of PbS.
Tnen do the same for 10 g Pb and 3.00 g S. You will find that the same amount of PbS is formed even though more S was used.
Then do the same for 18 g Pb. You will find the amount PbS is the same even though more Pb was used.
You can put all of that into your own words to explain the limiting reagent. Its Pb when more S is used but S when more Pb is used.
The concept of limiting reactant states that in a chemical reaction, the reactant that is completely consumed or used up first determines the maximum amount of product that can be formed. The other reactant, which is not completely used up, is called the excess reactant.
Let's analyze each situation:
1. 10.0 g of lead reacted with 1.56 g of sulfur:
To determine the limiting reactant, we can compare the moles of each reactant by dividing their masses by their molar masses (207.2 g/mol for lead and 32.1 g/mol for sulfur):
- The moles of lead = 10.0 g / 207.2 g/mol = 0.0482 mol
- The moles of sulfur = 1.56 g / 32.1 g/mol = 0.0485 mol
Since the moles of lead and sulfur are similar, we can conclude that the limiting reactant is sulfur. Therefore, all 1.56 g of sulfur react completely with a certain amount of lead to produce 11.56 g of lead sulfide.
2. 10.0 g of lead reacted with 3.00 g of sulfur:
Calculating the moles:
- Moles of lead = 10.0 g / 207.2 g/mol = 0.0482 mol
- Moles of sulfur = 3.00 g / 32.1 g/mol = 0.0935 mol
In this case, we can see that there is more sulfur than lead. Consequently, lead is the limiting reactant since it will be completely consumed first. Therefore, only 0.0482 mol of sulfur (1.56 g) will react with lead to produce 11.56 g of lead sulfide. The remaining excess sulfur (0.0935 mol - 0.0482 mol = 0.0453 mol) will not react.
3. 18.0 g of lead reacted with 1.56 g of sulfur:
Calculating the moles:
- Moles of lead = 18.0 g / 207.2 g/mol = 0.0868 mol
- Moles of sulfur = 1.56 g / 32.1 g/mol = 0.0485 mol
Comparing the moles, we can observe that there is an excess of lead. Therefore, sulfur is the limiting reactant. Only 0.0485 mol of sulfur (1.56 g) will react with the lead to produce 11.56 g of lead sulfide. The remaining excess lead (0.0868 mol - 0.0485 mol = 0.0383 mol) will not react.
In summary, the observations can be explained by the concept of limiting reactant. In every scenario, the maximum amount of lead sulfide formed is always 11.56 g, which corresponds to the complete conversion of the limiting reactant. The excess reactant does not affect the amount of product formed and remains unreacted.
To explain these observations in terms of the concept of limiting reactant, we need to understand what a limiting reactant is. In any chemical reaction, the reactants are typically present in specific quantities, and the reaction will proceed until one of the reactants is consumed completely. This reactant is known as the limiting reactant because it limits the amount of product that can be formed.
In the given scenarios, we have different quantities of lead and sulfur reacting to form lead sulfide. Let's analyze each situation individually:
1. 10.0 g of lead reacted with 1.56 g of sulfur:
In this case, we can calculate the amount of lead sulfide formed by determining the amount of reactant that can be completely consumed. Since lead is the limiting reactant, only 10.0 g of lead will react with 1.56 g of sulfur. The balanced chemical equation is:
2Pb + S -> 2PbS
To find the amount of lead sulfide formed, we need to compare the reactants' ratios in the balanced chemical equation. The molar ratio of lead to lead sulfide is 2:2 or 1:1. Therefore, the amount of lead sulfide formed is also 10.0 g.
2. 10.0 g of lead reacted with 3.00 g of sulfur:
In this scenario, we again compare the quantities of reactants. The molar ratio of lead to sulfur in the balanced chemical equation is 2:1. However, we have more sulfur (3.00 g) than needed to completely react with 10.0 g of lead. Therefore, 10.0 g of lead will be the limiting reactant, and the amount of lead sulfide formed will still be 10.0 g.
3. 18.0 g of lead reacted with 1.56 g of sulfur:
Once more, we compare the quantities of reactants. The molar ratio of lead to sulfur in the balanced chemical equation is still 2:1. Since we have a small amount of sulfur (1.56 g), it will be the limiting reactant, and only 1.56 g of lead will be consumed. Therefore, the amount of lead sulfide formed will be limited by the sulfur and will be less than 1.56 g.
In summary, the concept of limiting reactant helps us determine the maximum amount of product that can be formed in a chemical reaction. In these scenarios, the limiting reactant was either lead or sulfur, depending on their respective quantities. This concept allows us to explain why the amount of lead sulfide formed differs in each case.