How many grams of carbon dioxide is set free by the decomposition of 20 g of calcium carbonate?

Explanation

To solve the question of how many grams of carbon dioxide (CO₂) are produced from the decomposition of 20 grams of calcium carbonate (CaCO₃), we can follow these steps:

### Step 1: Write the decomposition reaction
The decomposition of calcium carbonate can be represented by the following chemical equation:
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]

### Step 2: Determine the molar mass of calcium carbonate (CaCO₃)
The molar mass of calcium carbonate can be calculated as follows:
- Calcium (Ca): 40.08 g/mol
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol (there are 3 oxygen atoms)

Calculating the molar mass:
\[ \text{Molar mass of CaCO}_3 = 40.08 + 12.01 + (3 \times 16.00) \]
\[ = 40.08 + 12.01 + 48.00 \]
\[ = 100.09 \, \text{g/mol} \]

### Step 3: Calculate the number of moles of calcium carbonate in 20 grams
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
We can find the number of moles of CaCO₃:
\[ \text{Number of moles of CaCO}_3 = \frac{20 \, \text{g}}{100.09 \, \text{g/mol}} \]
\[ \approx 0.1998 \, \text{mol} \]

### Step 4: Use the stoichiometry of the reaction
From the balanced equation, we see that 1 mole of CaCO₃ produces 1 mole of CO₂. Therefore, the number of moles of CO₂ produced will be the same as the number of moles of CaCO₃ decomposed:
\[ \text{Number of moles of CO}_2 = 0.1998 \, \text{mol} \]

### Step 5: Calculate the mass of carbon dioxide produced
Now, we need to calculate the mass of CO₂ produced using its molar mass:
- Molar mass of CO₂ = 12.01 g/mol (C) + 2 × 16.00 g/mol (O) = 44.01 g/mol

Using the formula:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]
We can find the mass of CO₂:
\[ \text{Mass of CO}_2 = 0.1998 \, \text{mol} \times 44.01 \, \text{g/mol} \]
\[ \approx 8.792 \, \text{g} \]

### Final Answer
Thus, the mass of carbon dioxide set free by the decomposition of 20 grams of calcium carbonate is approximately **8.79 grams**.

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