Consider the following reaction:2A  +  B  -----  3C  +  DIf 3.0 mol A and 2.0 mol B react to form 4.0 mol C, what is the percent yield of this reaction?

1. Break Down the Problem

To calculate the percent yield, we need to determine:

• The theoretical yield of product $C$ based on the available quantities of reactants $A$ and $B$.
• The actual yield of product $C$ which is given as 4.0 mol.
• Finally, calculate the percent yield using the formula:

$\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100$

2. Relevant Concepts

From the balanced reaction:

$2A + B \rightarrow 3C + D$

We see that:

• 2 moles of $A$ produce 3 moles of $C$.
• 1 mole of $B$ also contributes to the production of $C$.

3. Analysis and Detail

First, we need to determine the limiting reactant, which will inform us of the theoretical yield.

• Find moles of $C$ produced from $A$: $\text{From 3.0 mol of } A: \quad \frac{3 \text{ mol C}}{2 \text{ mol A}} \times 3.0 \text{ mol A} = 4.5 \text{ mol C}$

• Find moles of $C$ produced from $B$: $\text{From 2.0 mol of } B: \quad \frac{3 \text{ mol C}}{1 \text{ mol B}} \times 2.0 \text{ mol B} = 6.0 \text{ mol C}$

Since $A$ produces less $C$ (4.5 mol) than $B$ (6.0 mol), $A$ is the limiting reactant.

• Theoretical yield of $C$:
  The theoretical yield of $C$ when $A$ is the limiting reagent is 4.5 mol.

4. Verify and Summarize

• Actual Yield of $C$ is given as 4.0 mol.
• Theoretical Yield from $A$ is calculated to be 4.5 mol.

Now we can calculate the percent yield:

$\text{Percent Yield} = \left( \frac{4.0 \text{ mol}}{4.5 \text{ mol}} \right) \times 100 \approx 88.89 \%$

Final Answer

The percent yield of the reaction is approximately $88.89\%$.