10 g `CaCO_3` gives on strong heating `CO_2.` It gives quicklime ( in grams)

Options

A. 5 g
B. 4.4 g
C. 5.6 g
D. 4 g

Answer

To solve the problem of how much quicklime (CaO) is produced from heating 10 g of calcium carbonate (CaCO₃), we can follow these steps:

### Step 1: Write the decomposition reaction
When calcium carbonate is heated, it decomposes into calcium oxide (quicklime) and carbon dioxide (CO₂). The balanced chemical equation for this reaction is:
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]

### Step 2: Determine the molar masses
Next, we need to calculate the molar masses of CaCO₃ and CaO.

- **Molar mass of CaCO₃:**
- Calcium (Ca): 40 g/mol
- Carbon (C): 12 g/mol
- Oxygen (O): 16 g/mol (3 atoms)

\[ \text{Molar mass of CaCO}_3 = 40 + 12 + (16 \times 3) = 40 + 12 + 48 = 100 \text{ g/mol} \]

- **Molar mass of CaO:**
- Calcium (Ca): 40 g/mol
- Oxygen (O): 16 g/mol

\[ \text{Molar mass of CaO} = 40 + 16 = 56 \text{ g/mol} \]

### Step 3: Establish the stoichiometric relationship
From the balanced equation, we see that:
- 1 mole of CaCO₃ produces 1 mole of CaO.

### Step 4: Calculate the amount of CaO produced from 10 g of CaCO₃
Using the molar masses, we can find out how much CaO is produced from 10 g of CaCO₃.

1. First, calculate the number of moles of CaCO₃ in 10 g:
\[ \text{Moles of CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{10 \text{ g}}{100 \text{ g/mol}} = 0.1 \text{ mol} \]

2. Since the reaction shows that 1 mole of CaCO₃ produces 1 mole of CaO, 0.1 moles of CaCO₃ will produce 0.1 moles of CaO.

3. Now, calculate the mass of CaO produced:
\[ \text{Mass of CaO} = \text{moles} \times \text{molar mass} = 0.1 \text{ mol} \times 56 \text{ g/mol} = 5.6 \text{ g} \]

### Final Answer
The amount of quicklime (CaO) produced from heating 10 g of CaCO₃ is **5.6 grams**.

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