Understanding the Reaction
* Lead (Pb) and Sulfur (S) react to form Lead Sulfide (PbS). This is a classic example of a combination reaction.
* The Law of Conservation of Mass: This law states that matter cannot be created or destroyed in a chemical reaction. The total mass of the reactants must equal the total mass of the products.
Analyzing the Reaction Mixture
1. Limiting Reactant: The reactant that gets completely used up in a reaction is called the limiting reactant. It dictates how much product can be formed. To find this, we need to calculate the moles of each reactant.
* Moles of Pb = (10.00 g Pb) / (207.2 g/mol Pb) = 0.0483 mol Pb
* Moles of S = (1.56 g S) / (32.07 g/mol S) = 0.0487 mol S
Since the stoichiometry of the reaction is 1:1 (1 mole of Pb reacts with 1 mole of S), the limiting reactant is lead (Pb).
2. Theoretical Yield: This is the maximum amount of product that can be formed based on the limiting reactant.
* Moles of PbS = 0.0483 mol Pb (since 1 mol Pb produces 1 mol PbS)
* Theoretical yield of PbS = (0.0483 mol PbS) * (239.3 g/mol PbS) = 11.54 g PbS
3. Actual Yield: This is the amount of product that is actually obtained from the experiment. We are given an actual yield of 11.56 g PbS.
4. Excess Reactant: The reactant that is not completely used up is the excess reactant. In this case, it's sulfur (S).
5. Reaction Mixture after 30.00 g of Lead:
* We started with 10.00 g of lead and added 30.00 g more, giving us a total of 40.00 g of lead.
* Since we have a large excess of lead, the limiting reactant will still be sulfur (S).
* The reaction will consume all 1.56 g of sulfur, producing 11.54 g of lead sulfide.
Therefore, the reaction mixture after adding 30.00 g of lead would contain:
* 11.54 g of Lead Sulfide (PbS) (the maximum amount that can be formed)
* 28.46 g of Lead (Pb) (excess reactant)
* No Sulfur (S) (completely consumed)
Important Note: The theoretical yield and actual yield are very close, suggesting a very efficient reaction.
