Answer :
Certainly! Let's solve this problem step by step.
### Step 1: Balance the Chemical Equation
The decomposition reaction of calcium carbonate (CaCO₃) is:
[tex]\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \][/tex]
This equation is already balanced, as there is one calcium atom, one carbon atom, and three oxygen atoms on both sides.
### Step 2: Calculate Molar Masses
We need the molar masses of the compounds involved:
- Calcium Carbonate (CaCO₃):
- Calcium (Ca): 40.08 g/mol
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol (3 atoms, so 3 16.00)
- Total: 40.08 + 12.01 + 48.00 = 100.09 g/mol
- Carbon Dioxide (CO₂):
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol (2 atoms, so 2 16.00)
- Total: 12.01 + 32.00 = 44.01 g/mol
### Step 3: Calculate Moles of CaCO₃
Given that you start with 235.0 g of CaCO₃, calculate the number of moles:
[tex]\[
\text{Moles of CaCO}_3 = \frac{235.0 \, \text{g}}{100.09 \, \text{g/mol}} \approx 2.3479 \, \text{moles}
\][/tex]
### Step 4: Calculate Theoretical Mass of CO₂
From the balanced equation, 1 mole of CaCO₃ produces 1 mole of CO₂. Thus, the moles of CO₂ produced should equal the moles of CaCO₃:
[tex]\[
\text{Theoretical mass of CO}_2 = \text{Moles of CO}_2 \times \text{Molar mass of CO}_2 = 2.3479 \, \text{moles} \times 44.01 \, \text{g/mol} \approx 103.33 \, \text{g}
\][/tex]
### Step 5: Calculate Percent Yield
Actual mass of CO₂ produced = 97.5 g
To find the percent yield:
[tex]\[
\text{Percent yield} = \left( \frac{\text{Actual mass of CO}_2}{\text{Theoretical mass of CO}_2} \right) \times 100
\][/tex]
[tex]\[
\text{Percent yield} = \left( \frac{97.5 \, \text{g}}{103.33 \, \text{g}} \right) \times 100 \approx 94.36\%
\][/tex]
So, the percent yield of CO₂ in this reaction is approximately 94.36%.
### Step 1: Balance the Chemical Equation
The decomposition reaction of calcium carbonate (CaCO₃) is:
[tex]\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \][/tex]
This equation is already balanced, as there is one calcium atom, one carbon atom, and three oxygen atoms on both sides.
### Step 2: Calculate Molar Masses
We need the molar masses of the compounds involved:
- Calcium Carbonate (CaCO₃):
- Calcium (Ca): 40.08 g/mol
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol (3 atoms, so 3 16.00)
- Total: 40.08 + 12.01 + 48.00 = 100.09 g/mol
- Carbon Dioxide (CO₂):
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol (2 atoms, so 2 16.00)
- Total: 12.01 + 32.00 = 44.01 g/mol
### Step 3: Calculate Moles of CaCO₃
Given that you start with 235.0 g of CaCO₃, calculate the number of moles:
[tex]\[
\text{Moles of CaCO}_3 = \frac{235.0 \, \text{g}}{100.09 \, \text{g/mol}} \approx 2.3479 \, \text{moles}
\][/tex]
### Step 4: Calculate Theoretical Mass of CO₂
From the balanced equation, 1 mole of CaCO₃ produces 1 mole of CO₂. Thus, the moles of CO₂ produced should equal the moles of CaCO₃:
[tex]\[
\text{Theoretical mass of CO}_2 = \text{Moles of CO}_2 \times \text{Molar mass of CO}_2 = 2.3479 \, \text{moles} \times 44.01 \, \text{g/mol} \approx 103.33 \, \text{g}
\][/tex]
### Step 5: Calculate Percent Yield
Actual mass of CO₂ produced = 97.5 g
To find the percent yield:
[tex]\[
\text{Percent yield} = \left( \frac{\text{Actual mass of CO}_2}{\text{Theoretical mass of CO}_2} \right) \times 100
\][/tex]
[tex]\[
\text{Percent yield} = \left( \frac{97.5 \, \text{g}}{103.33 \, \text{g}} \right) \times 100 \approx 94.36\%
\][/tex]
So, the percent yield of CO₂ in this reaction is approximately 94.36%.
