Question
A scientist heats 30 grams of calcium carbonate in the chemical reaction shown CaCO3-CO2+CaO
Answers
GPT-4o mini
When calcium carbonate (CaCO₃) is heated, it undergoes a thermal decomposition reaction, breaking down into calcium oxide (CaO) and carbon dioxide (CO₂). The balanced chemical equation for this reaction is:
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]
To analyze the reaction further, let's consider the heating of 30 grams of calcium carbonate.
### Step 1: Determine the molar mass of calcium carbonate.
- Calcium (Ca): approximately 40.08 g/mol
- Carbon (C): approximately 12.01 g/mol
- Oxygen (O): approximately 16.00 g/mol (and there are three of them in CaCO₃)
Calculating the molar mass:
\[
\text{Molar mass of CaCO}_3 = 40.08 + 12.01 + (3 \times 16.00) = 100.09 \, \text{g/mol}
\]
### Step 2: Calculate the number of moles of calcium carbonate in 30 grams.
Using the formula:
\[
\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}
\]
\[
\text{moles of CaCO}_3 = \frac{30 \, \text{g}}{100.09 \, \text{g/mol}} \approx 0.2997 \, \text{mol}
\]
### Step 3: Calculate the amount of products formed.
From the balanced equation, we see that 1 mole of CaCO₃ produces 1 mole of CaO and 1 mole of CO₂. Thus, the moles of products formed are the same as the moles of reactants.
- Moles of CaO produced = 0.2997 mol
- Moles of CO₂ produced = 0.2997 mol
### Step 4: Convert moles of products to grams (if desired).
- **Molar mass of CaO**:
- Ca: 40.08 g/mol
- O: 16.00 g/mol
\[
\text{Molar mass of CaO} = 40.08 + 16.00 = 56.08 \, \text{g/mol}
\]
- **Molar mass of CO₂**:
- C: 12.01 g/mol
- O: 16.00 g/mol (and there are two of them in CO₂)
\[
\text{Molar mass of CO}_2 = 12.01 + (2 \times 16.00) = 44.01 \, \text{g/mol}
\]
- Mass of CaO produced:
\[
\text{mass of CaO} = 0.2997 \, \text{mol} \times 56.08 \, \text{g/mol} \approx 16.81 \, \text{g}
\]
- Mass of CO₂ produced:
\[
\text{mass of CO}_2 = 0.2997 \, \text{mol} \times 44.01 \, \text{g/mol} \approx 13.19 \, \text{g}
\]
### Summary:
When 30 grams of calcium carbonate is heated, it decomposes to produce approximately:
- 16.81 grams of calcium oxide (CaO)
- 13.19 grams of carbon dioxide (CO₂)
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]
To analyze the reaction further, let's consider the heating of 30 grams of calcium carbonate.
### Step 1: Determine the molar mass of calcium carbonate.
- Calcium (Ca): approximately 40.08 g/mol
- Carbon (C): approximately 12.01 g/mol
- Oxygen (O): approximately 16.00 g/mol (and there are three of them in CaCO₃)
Calculating the molar mass:
\[
\text{Molar mass of CaCO}_3 = 40.08 + 12.01 + (3 \times 16.00) = 100.09 \, \text{g/mol}
\]
### Step 2: Calculate the number of moles of calcium carbonate in 30 grams.
Using the formula:
\[
\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}
\]
\[
\text{moles of CaCO}_3 = \frac{30 \, \text{g}}{100.09 \, \text{g/mol}} \approx 0.2997 \, \text{mol}
\]
### Step 3: Calculate the amount of products formed.
From the balanced equation, we see that 1 mole of CaCO₃ produces 1 mole of CaO and 1 mole of CO₂. Thus, the moles of products formed are the same as the moles of reactants.
- Moles of CaO produced = 0.2997 mol
- Moles of CO₂ produced = 0.2997 mol
### Step 4: Convert moles of products to grams (if desired).
- **Molar mass of CaO**:
- Ca: 40.08 g/mol
- O: 16.00 g/mol
\[
\text{Molar mass of CaO} = 40.08 + 16.00 = 56.08 \, \text{g/mol}
\]
- **Molar mass of CO₂**:
- C: 12.01 g/mol
- O: 16.00 g/mol (and there are two of them in CO₂)
\[
\text{Molar mass of CO}_2 = 12.01 + (2 \times 16.00) = 44.01 \, \text{g/mol}
\]
- Mass of CaO produced:
\[
\text{mass of CaO} = 0.2997 \, \text{mol} \times 56.08 \, \text{g/mol} \approx 16.81 \, \text{g}
\]
- Mass of CO₂ produced:
\[
\text{mass of CO}_2 = 0.2997 \, \text{mol} \times 44.01 \, \text{g/mol} \approx 13.19 \, \text{g}
\]
### Summary:
When 30 grams of calcium carbonate is heated, it decomposes to produce approximately:
- 16.81 grams of calcium oxide (CaO)
- 13.19 grams of carbon dioxide (CO₂)