20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What be the percentage purity of magnsesium carbonate in the sample?

Options

A. 75
B. 96
C. 60
D. 84

Explanation

To find the percentage purity of magnesium carbonate (MgCO3) in the sample, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the magnesium carbonate sample = 20.0 g
- Mass of magnesium oxide (MgO) produced = 8.0 g

2. **Write the Decomposition Reaction:**
The thermal decomposition of magnesium carbonate can be represented as:
\[ \text{MgCO}_3 \rightarrow \text{MgO} + \text{CO}_2 \]
From the reaction, we can see that 1 mole of MgCO3 produces 1 mole of MgO.

3. **Calculate the Molar Mass of MgO:**
- Molar mass of Mg = 24 g/mol
- Molar mass of O = 16 g/mol
- Therefore, molar mass of MgO = 24 + 16 = 40 g/mol

4. **Calculate the Number of Moles of MgO Produced:**
\[ \text{Number of moles of MgO} = \frac{\text{mass of MgO}}{\text{molar mass of MgO}} = \frac{8.0 \text{ g}}{40 \text{ g/mol}} = 0.2 \text{ moles} \]

5. **Determine the Number of Moles of MgCO3:**
Since the stoichiometry of the reaction shows that 1 mole of MgCO3 produces 1 mole of MgO, the number of moles of MgCO3 is also 0.2 moles.

6. **Calculate the Mass of MgCO3:**
- Molar mass of MgCO3:
- Mg = 24 g/mol
- C = 12 g/mol
- O = 16 g/mol (3 oxygen atoms)
- Therefore, molar mass of MgCO3 = 24 + 12 + (3 × 16) = 84 g/mol
\[ \text{Mass of MgCO3} = \text{Number of moles} \times \text{Molar mass} = 0.2 \text{ moles} \times 84 \text{ g/mol} = 16.8 \text{ g} \]

7. **Calculate the Percentage Purity of MgCO3:**
\[ \text{Percentage purity} = \left( \frac{\text{mass of pure MgCO3}}{\text{mass of sample}} \right) \times 100 = \left( \frac{16.8 \text{ g}}{20.0 \text{ g}} \right) \times 100 = 84\% \]

### Final Answer:
The percentage purity of magnesium carbonate in the sample is **84%**.