What is correct for 10 g of `CaCO_(3)`?

Options

A. It contains 1 g atom of carbon
B. It contains `0*3` g atoms of oxygen
C. It contains 12 g of calcium
D. It refers to `0*1` g-equivalent of `CaCO_(3)`

Explanation

To solve the question regarding the composition of 10 g of \( \text{CaCO}_3 \), we will follow these steps:

### Step 1: Calculate the Molar Mass of \( \text{CaCO}_3 \)

The molar mass of \( \text{CaCO}_3 \) can be calculated by adding the atomic masses of its constituent elements:
- Calcium (Ca): 40 g/mol
- Carbon (C): 12 g/mol
- Oxygen (O): 16 g/mol (and there are 3 oxygen atoms)

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

### Step 2: Calculate the Number of Moles of \( \text{CaCO}_3 \)

To find the number of moles of \( \text{CaCO}_3 \) in 10 g, we use the formula:

\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{10 \text{ g}}{100 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 3: Calculate the Number of Gram Atoms of Carbon

Since there is 1 atom of carbon in each molecule of \( \text{CaCO}_3 \):

\[ \text{Gram atoms of Carbon} = \text{Number of moles} \times 1 = 0.1 \text{ moles} \times 1 = 0.1 \text{ gram atoms} \]

### Step 4: Calculate the Number of Gram Atoms of Oxygen

There are 3 oxygen atoms in each molecule of \( \text{CaCO}_3 \):

\[ \text{Gram atoms of Oxygen} = \text{Number of moles} \times 3 = 0.1 \text{ moles} \times 3 = 0.3 \text{ gram atoms} \]

### Step 5: Calculate the Mass of Calcium in 10 g of \( \text{CaCO}_3 \)

The mass of calcium in \( \text{CaCO}_3 \) can be calculated using the ratio of the molar mass of calcium to the molar mass of \( \text{CaCO}_3 \):

\[ \text{Mass of Calcium} = \left( \frac{40 \text{ g}}{100 \text{ g}} \right) \times 10 \text{ g} = 4 \text{ g} \]

### Step 6: Calculate the Gram Equivalent of \( \text{CaCO}_3 \)

The gram equivalent is calculated using the formula:

\[ \text{Gram equivalent} = \frac{\text{Number of moles}}{\text{n}} \quad \text{(where n = number of replaceable hydrogen ions or valency)} \]

For \( \text{CaCO}_3 \), n = 2 (since it can replace 2 H ions in reactions):

\[ \text{Gram equivalent} = 0.1 \text{ moles} \times 2 = 0.2 \text{ gram equivalents} \]

### Conclusion

From the calculations, we find:
- It contains 0.1 gram atoms of carbon.
- It contains 0.3 gram atoms of oxygen.
- It contains 4 g of calcium.
- It refers to 0.2 gram equivalents of \( \text{CaCO}_3 \).

Thus, the correct statement for 10 g of \( \text{CaCO}_3 \) is that it contains **0.3 gram atoms of oxygen**.