J. J. Berzelius collected the following data for three reactions:
Reaction 1 – 11.56 grams of lead sulfide, PbS, were formed when 10.0 grams of lead reacted with 1.56 grams of sulfur
Reaction 2 – 11.56 grams of lead sulfide were formed when 10.0 grams of lead reacted with 3.00 grams of sulfur
Reaction 3 – 11.56 grams of lead sulfide were formed when 18.0 grams of lead reacted with 1.56 grams of sulfur. Explain his observations in terms of the concept of limiting reactants.
Pb + S ==> PbS
10.0..1.56..11.56
10.0..3.00..11.56
18.0..1.56..11.56
10.0g Pb and 1.56 g S are the stoichiometric quantities for PbS to form. Anything more than 10.0 g Pb/1.56 g S will not produce anymore PbS. That is, 10.0 g Pb and more S than 1.56 still produces 11.56g PbS and 1.56 g S and more Pb than 10.0 g will still produce only 11.56g PbS.
To explain J.J. Berzelius' observations in terms of the concept of limiting reactants, we need to understand what a limiting reactant is and how it affects the amount of product formed in a chemical reaction.
In a chemical reaction, reactants combine to form products. The reactant that is completely consumed or used up first is called the limiting reactant. This reactant limits the amount of product that can be formed because once it is used up, the reaction can no longer proceed.
Let's analyze the three reactions mentioned:
Reaction 1: 10.0 grams of lead (Pb) reacts with 1.56 grams of sulfur (S) to form 11.56 grams of lead sulfide (PbS).
Reaction 2: 10.0 grams of lead (Pb) reacts with 3.00 grams of sulfur (S) to form 11.56 grams of lead sulfide (PbS).
Reaction 3: 18.0 grams of lead (Pb) reacts with 1.56 grams of sulfur (S) to form 11.56 grams of lead sulfide (PbS).
In all three reactions, the amount of lead sulfide formed is the same (11.56 grams). However, the amounts of lead (Pb) and sulfur (S) used vary.
By comparing the amounts of lead (Pb) and sulfur (S) in each reaction, we can identify the limiting reactant.
Let's consider Reaction 1:
Lead (Pb) = 10.0 grams
Sulfur (S) = 1.56 grams
Using the balanced chemical equation:
2Pb + S → PbS
From the equation, we can see that the mole ratio between lead (Pb) and sulfur (S) is 2:1. Thus, 2 moles of lead react with 1 mole of sulfur to form 1 mole of lead sulfide.
To find the limiting reactant, we need to convert the mass of each reactant to moles.
Molar mass of Pb = 207.2 g/mol
Molar mass of S = 32.1 g/mol
Moles of lead (Pb) = 10.0 g Pb × (1 mol Pb / 207.2 g Pb) = 0.048 mol Pb
Moles of sulfur (S) = 1.56 g S × (1 mol S / 32.1 g S) = 0.048 mol S
Since the moles of lead (Pb) and sulfur (S) are the same in Reaction 1, we can conclude that neither reactant is in excess, and both are fully consumed. Hence, the stoichiometry of the reaction is satisfied.
Now, let's consider Reaction 2:
Lead (Pb) = 10.0 grams
Sulfur (S) = 3.00 grams
Moles of lead (Pb) = 10.0 g Pb × (1 mol Pb / 207.2 g Pb) = 0.048 mol Pb
Moles of sulfur (S) = 3.00 g S × (1 mol S / 32.1 g S) = 0.093 mol S
From the balanced equation, we know that 2 moles of lead react with 1 mole of sulfur to form 1 mole of lead sulfide.
Comparing the moles of lead (Pb) and sulfur (S), we see that the mole ratio is 2:1, indicating that sulfur is present in excess.
Since sulfur is in excess, the amount of lead sulfide formed will still be determined by the limited amount of lead (Pb). Therefore, Reaction 2 produces the same amount of lead sulfide as Reaction 1.
Lastly, let's consider Reaction 3:
Lead (Pb) = 18.0 grams
Sulfur (S) = 1.56 grams
Moles of lead (Pb) = 18.0 g Pb × (1 mol Pb / 207.2 g Pb) = 0.087 mol Pb
Moles of sulfur (S) = 1.56 g S × (1 mol S / 32.1 g S) = 0.048 mol S
From the balanced equation, we know that 2 moles of lead react with 1 mole of sulfur to form 1 mole of lead sulfide.
Comparing the moles of lead (Pb) and sulfur (S), we see that the mole ratio is 2:1, indicating that lead is present in excess.
Since lead is in excess, the amount of lead sulfide formed will still be determined by the limited amount of sulfur (S). Therefore, Reaction 3 produces the same amount of lead sulfide as Reaction 1.
Based on these observations, we can conclude that the concept of the limiting reactant explains Berzelius' observations. In all three reactions, the amount of lead sulfide formed is the same because the limiting reactant, either lead or sulfur, is fully consumed and determines the amount of product formed.