Question
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.
Calculate the theoretical yield of C.
Round your answer to the nearest 0.1 g.
Answers
GPT-4o
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.
\[ A + B \rightarrow C \]
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.