Based on the following reaction 2A + 3B --> A2B3, if 1 mole of A is reacted with 2 moles of B, explain how you would determine which of the two substances is the limiting reactant.
To determine the limiting reactant, we need to compare the given moles of each reactant to the stoichiometry of the balanced chemical equation. The stoichiometry tells us the mole ratio between the reactants and products.
In this case, the balanced chemical equation is:
2A + 3B --> A2B3
Given:
- 1 mole of A
- 2 moles of B
First, we need to check how many moles of A2B3 can be formed from the given amounts of A and B.
Using the stoichiometry from the balanced equation, we can see that for every 2 moles of A reacted, we produce 1 mole of A2B3. Similarly, for every 3 moles of B reacted, we produce 1 mole of A2B3.
Now, we compare the amounts of A and B with the stoichiometry:
- Moles of A2B3 from A:
1 mole of A can produce 1/2 mole of A2B3 (using the stoichiometry ratio of 2:1)
- Moles of A2B3 from B:
2 moles of B can produce 2/3 mole of A2B3 (using the stoichiometry ratio of 3:1)
Since we have less moles of A2B3 from A compared to B, the limiting reactant is A. This means that A will be completely consumed in the reaction, while B will still be leftover.
Therefore, A is the limiting reactant in this reaction.
To determine the limiting reactant in a chemical reaction, you need to compare the molar ratios of the reactants to the stoichiometric ratio in the balanced equation.
In the given reaction, the balanced equation shows that 2 moles of A react with 3 moles of B to form 1 mole of A2B3. From this information, you can compute the ratio of moles of A to moles of B:
A:B = 2:3
Now, let's compare this ratio with the actual amounts provided in the question, which is 1 mole of A and 2 moles of B.
For A, the ratio can be calculated as:
(1 mole of A) / (2 moles of B) = 0.5
For B, the ratio can be calculated as:
(2 moles of B) / (1 mole of A) = 2
Since the ratio of A is 0.5 and B is 2, we can conclude that the limiting reactant is the one with the smaller ratio. In this case, A has the smaller ratio, indicating that A is the limiting reactant.
Therefore, in the given reaction, if 1 mole of A is reacted with 2 moles of B, A will be the limiting reactant since there is an insufficient amount of A to fully react with the available 2 moles of B.
2A + 3B ==> A2B3
1 mol A x (1 mol A2B3/2mols A) = 1 x 1/2 = 1/2 mol A2B3 formed.
2mols B x (1 mol A2B3/3mols B) = 2 x 1/3 = 2/3 mol A2B3 formed.
In limiting reagent problems, the SMALLER number is ALWAYS the correct value and the reagent producing that number is the limiting reagent. So which is smaller, 1/2 or 2/3.