To calculate the theoretical yield of Compound C, we need to use a balanced chemical equation and the molar masses of the compounds involved. However, since the actual equation and molar masses are not provided, I'll assume that the balanced equation is:
\[ A + B \rightarrow C \]
Let's also assume that you have the molar masses of compounds \( A \) and \( B \). Here is the general procedure to find the theoretical yield of \( C \):
1. **Calculate the moles of \( A \) and \( B \):**
Using the molar masses, calculate the moles of each reactant.
\[
\text{Moles of } A = \frac{\text{Mass of } A}{\text{Molar Mass of } A}
\]
\[
\text{Moles of } B = \frac{\text{Mass of } B}{\text{Molar Mass of } B}
\]
2. **Identify the limiting reagent:**
Determine which reactant is the limiting reagent (the reactant that will be completely consumed first and thus limits the amount of product formed).
3. **Calculate the moles of \( C \) formed:**
Use the stoichiometry of the balanced equation to find the moles of \( C \) produced from the limiting reagent.
4. **Calculate the mass of \( C \):**
Convert the moles of \( C \) to grams.
\[
\text{Mass of } C = \text{Moles of } C \times \text{Molar Mass of } C
\]
Since we don't have actual molar masses here, I'll provide a general example:
### Example with Hypothetical Molar Masses
Let's assume the molar masses are:
- Molar mass of \( A \) = 50 g/mol
- Molar mass of \( B \) = 110 g/mol
- Molar mass of \( C \) = 160 g/mol
Step 1: Calculate moles of \( A \) and \( B \):
\[ \text{Moles of } A = \frac{1.0 \text{ g }}{50 \text{ g/mol}} = 0.02 \text{ mol} \]
\[ \text{Moles of } B = \frac{5.5 \text{ g }}{110 \text{ g/mol}} = 0.05 \text{ mol} \]
Step 2: Identify the limiting reagent:
From the hypothetical balanced equation \( A + B \rightarrow C \), the stoichiometry is 1:1:1. Therefore, 0.02 mol of \( A \) would require 0.02 mol of \( B \). Since we have 0.05 mol of \( B \), \( A \) is the limiting reagent.
Step 3: Calculate the moles of \( C \) produced:
Since the ratio is 1:1, the moles of \( C \) produced will be the same as the moles of the limiting reagent, \( A \):
\[ \text{Moles of } C = 0.02 \text{ mol} \]
Step 4: Calculate the mass of \( C \):
\[ \text{Mass of } C = 0.02 \text{ mol} \times 160 \text{ g/mol} = 3.2 \text{ g} \]
Therefore, the theoretical yield of \( C \) is 3.2 g.
Suppose 1.0 g of Compound A are consumed in a reaction with 5.5 g of Compound B that produces only one product, Compound C.
Calculate the theoretical yield of C.
Round your answer to the nearest 0.1 g.
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