Sodium carbonate of 92% purity is used in the reation `Na_(2)CO_(3)+CaCl_(2)toCaCO_(3)+2NaCl`. The number of grams of `Na_(2)CO_(3)` required to yield 1 gm of `CaCO_(3)`

Options

A. `8.5 g`
B. `10.5 g`
C. `11.52 g`
D. `1.152 g`

Answer

To solve the problem of how many grams of sodium carbonate (Na₂CO₃) of 92% purity are required to yield 1 gram of calcium carbonate (CaCO₃), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ \text{Na}_2\text{CO}_3 + \text{CaCl}_2 \rightarrow \text{CaCO}_3 + 2\text{NaCl} \]

### Step 2: Determine the molar masses
- Molar mass of Na₂CO₃ (sodium carbonate):
- Na: 23 g/mol × 2 = 46 g/mol
- C: 12 g/mol × 1 = 12 g/mol
- O: 16 g/mol × 3 = 48 g/mol
- Total = 46 + 12 + 48 = 106 g/mol

- Molar mass of CaCO₃ (calcium carbonate):
- Ca: 40 g/mol × 1 = 40 g/mol
- C: 12 g/mol × 1 = 12 g/mol
- O: 16 g/mol × 3 = 48 g/mol
- Total = 40 + 12 + 48 = 100 g/mol

### Step 3: Calculate the amount of Na₂CO₃ needed for 1 g of CaCO₃
From the stoichiometry of the reaction, 1 mole of Na₂CO₃ produces 1 mole of CaCO₃. Therefore, the amount of Na₂CO₃ needed to produce 100 g of CaCO₃ is 106 g.

To find out how much Na₂CO₃ is needed to produce 1 g of CaCO₃:
\[ \text{Amount of Na}_2\text{CO}_3 = \frac{106 \text{ g Na}_2\text{CO}_3}{100 \text{ g CaCO}_3} \times 1 \text{ g CaCO}_3 = 1.06 \text{ g Na}_2\text{CO}_3 \]

### Step 4: Adjust for the purity of Na₂CO₃
Since the sodium carbonate is only 92% pure, we need to find the actual mass of Na₂CO₃ required to yield 1.06 g of pure Na₂CO₃.

Let \( x \) be the mass of the 92% pure Na₂CO₃ required:
\[ x \times \frac{92}{100} = 1.06 \]

### Step 5: Solve for \( x \)
Rearranging the equation gives:
\[ x = \frac{1.06 \times 100}{92} \]
\[ x = \frac{106}{92} \]
\[ x \approx 1.152 \text{ g} \]

### Final Answer
Thus, the number of grams of sodium carbonate required to yield 1 gram of calcium carbonate is approximately **1.152 grams**.

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