Acetylene can be prepared from calcium carbonate by a series of reactions. The mass of 80% calcium carbonate required to prepare 2 moles of acetylene is

Options

A. a. 200g
B. b. 160g
C. c. 250 g
D. d. 320 g

Answer

To solve the problem of how much 80% calcium carbonate (CaCO3) is required to prepare 2 moles of acetylene (C2H2), we will follow these steps:

### Step 1: Understand the Reaction Process
The preparation of acetylene from calcium carbonate involves a series of reactions:
1. **Decomposition of Calcium Carbonate**:
\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]
2. **Formation of Calcium Carbide**:
\[ \text{CaO} + 3\text{C} \rightarrow \text{CaC}_2 + \text{CO} \]
3. **Hydrolysis of Calcium Carbide**:
\[ \text{CaC}_2 + 2\text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_2 + \text{Ca(OH)}_2 \]

From these reactions, we can see that 1 mole of CaCO3 produces 1 mole of C2H2.

### Step 2: Calculate Moles of CaCO3 Required
Since we need to produce 2 moles of acetylene (C2H2), we will need 2 moles of calcium carbonate (CaCO3):
\[ \text{Moles of CaCO}_3 = 2 \text{ moles} \]

### Step 3: Calculate the Mass of Pure CaCO3 Required
The molar mass of calcium carbonate (CaCO3) is approximately:
- Ca: 40 g/mol
- C: 12 g/mol
- O3: 16 g/mol × 3 = 48 g/mol
- Total: 40 + 12 + 48 = 100 g/mol

Thus, the mass of 2 moles of CaCO3 is:
\[ \text{Mass of CaCO}_3 = 2 \text{ moles} \times 100 \text{ g/mol} = 200 \text{ grams} \]

### Step 4: Adjust for the Purity of Calcium Carbonate
Since we have 80% calcium carbonate, we need to find the mass of the 80% solution that contains 200 grams of pure CaCO3. Let \( x \) be the mass of the 80% CaCO3 required:
\[ 0.80x = 200 \text{ grams} \]
To find \( x \):
\[ x = \frac{200}{0.80} = 250 \text{ grams} \]

### Conclusion
The mass of 80% calcium carbonate required to prepare 2 moles of acetylene is **250 grams**.