Question
Calcium carbonate (CaCO3) can undergo a chemical reaction to produce calcium oxide (CaO) and carbon dioxide (CO2). Which possible masses of calcium oxide and carbon dioxide can form when 200 g of calcium carbonate undergoes this chemical reaction?(1 point)
Responses
56 g CaO and 44 g CO2subscript
56 g CaO and 44 g CO 2 subscript
112 g CaO and 88 g CO2subscript
112 g CaO and 88 g CO 2 subscript
200 g CaO and 200 g CO2subscript
200 g CaO and 200 g CO 2 subscript
40 g CaO and 12 g CO2
Responses
56 g CaO and 44 g CO2subscript
56 g CaO and 44 g CO 2 subscript
112 g CaO and 88 g CO2subscript
112 g CaO and 88 g CO 2 subscript
200 g CaO and 200 g CO2subscript
200 g CaO and 200 g CO 2 subscript
40 g CaO and 12 g CO2
Answers
GPT-4o mini
To determine the possible masses of calcium oxide (CaO) and carbon dioxide (CO2) produced from the decomposition of calcium carbonate (CaCO3), we first need to consider the balanced chemical equation for the reaction:
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]
From the balanced equation, we see that 1 mole of calcium carbonate decomposes to form 1 mole of calcium oxide and 1 mole of carbon dioxide.
**Step 1: Calculate the molar masses.**
- Molar mass of CaCO3 = 40.08 g/mol (Ca) + 12.01 g/mol (C) + 3 × 16.00 g/mol (O)
= 40.08 + 12.01 + 48.00
= 100.09 g/mol
- Molar mass of CaO = 40.08 g/mol (Ca) + 16.00 g/mol (O)
= 40.08 + 16.00
= 56.08 g/mol
- Molar mass of CO2 = 12.01 g/mol (C) + 2 × 16.00 g/mol (O)
= 12.01 + 32.00
= 44.01 g/mol
**Step 2: Determine moles of CaCO3 from 200 g.**
\[
\text{Moles of CaCO}_3 = \frac{200 \, \text{g}}{100.09 \, \text{g/mol}} \approx 1.996 \, \text{mol}
\]
**Step 3: Use the stoichiometry of the reaction.**
From the balanced equation:
- 1 mole of CaCO3 produces 1 mole of CaO and 1 mole of CO2.
Therefore:
- Moles of CaO produced = Moles of CaCO3 = 1.996 mol
- Moles of CO2 produced = Moles of CaCO3 = 1.996 mol
**Step 4: Convert moles of products to grams.**
- Mass of CaO produced:
\[
\text{Mass of CaO} = \text{Moles of CaO} \times \text{Molar mass of CaO} = 1.996 \, \text{mol} \times 56.08 \, \text{g/mol} \approx 112.0 \, \text{g}
\]
- Mass of CO2 produced:
\[
\text{Mass of CO}_2 = \text{Moles of CO}_2 \times \text{Molar mass of CO}_2 = 1.996 \, \text{mol} \times 44.01 \, \text{g/mol} \approx 88.0 \, \text{g}
\]
**Final Results:**
When 200 g of CaCO3 undergoes decomposition, it produces approximately:
- 112 g of CaO
- 88 g of CO2
Among the provided options, the correct response is:
**112 g CaO and 88 g CO2.**
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]
From the balanced equation, we see that 1 mole of calcium carbonate decomposes to form 1 mole of calcium oxide and 1 mole of carbon dioxide.
**Step 1: Calculate the molar masses.**
- Molar mass of CaCO3 = 40.08 g/mol (Ca) + 12.01 g/mol (C) + 3 × 16.00 g/mol (O)
= 40.08 + 12.01 + 48.00
= 100.09 g/mol
- Molar mass of CaO = 40.08 g/mol (Ca) + 16.00 g/mol (O)
= 40.08 + 16.00
= 56.08 g/mol
- Molar mass of CO2 = 12.01 g/mol (C) + 2 × 16.00 g/mol (O)
= 12.01 + 32.00
= 44.01 g/mol
**Step 2: Determine moles of CaCO3 from 200 g.**
\[
\text{Moles of CaCO}_3 = \frac{200 \, \text{g}}{100.09 \, \text{g/mol}} \approx 1.996 \, \text{mol}
\]
**Step 3: Use the stoichiometry of the reaction.**
From the balanced equation:
- 1 mole of CaCO3 produces 1 mole of CaO and 1 mole of CO2.
Therefore:
- Moles of CaO produced = Moles of CaCO3 = 1.996 mol
- Moles of CO2 produced = Moles of CaCO3 = 1.996 mol
**Step 4: Convert moles of products to grams.**
- Mass of CaO produced:
\[
\text{Mass of CaO} = \text{Moles of CaO} \times \text{Molar mass of CaO} = 1.996 \, \text{mol} \times 56.08 \, \text{g/mol} \approx 112.0 \, \text{g}
\]
- Mass of CO2 produced:
\[
\text{Mass of CO}_2 = \text{Moles of CO}_2 \times \text{Molar mass of CO}_2 = 1.996 \, \text{mol} \times 44.01 \, \text{g/mol} \approx 88.0 \, \text{g}
\]
**Final Results:**
When 200 g of CaCO3 undergoes decomposition, it produces approximately:
- 112 g of CaO
- 88 g of CO2
Among the provided options, the correct response is:
**112 g CaO and 88 g CO2.**