Calculate the number of hours of service that can be derived at 1 atm, 300 K from an acetylene lamp containing 640 g calcium carbide. Given that the lamp requires 50 L acetylene gas at 1 atm 300 K for one hour. `["Take "0.0821xx300=25]`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Write the reaction for the formation of acetylene from calcium carbide.
The reaction is:
\[ \text{CaC}_2 + \text{2H}_2O \rightarrow \text{C}_2\text{H}_2 + \text{Ca(OH)}_2 \]

### Step 2: Calculate the molar mass of calcium carbide (CaC2).
- Atomic weight of Calcium (Ca) = 40 g/mol
- Atomic weight of Carbon (C) = 12 g/mol
- Molar mass of CaC2 = 40 + (2 × 12) = 40 + 24 = 64 g/mol

### Step 3: Determine the number of moles of calcium carbide in 640 g.
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
\[ \text{Number of moles of CaC}_2 = \frac{640 \text{ g}}{64 \text{ g/mol}} = 10 \text{ moles} \]

### Step 4: Calculate the volume of acetylene gas produced using the ideal gas law.
Using the ideal gas equation:
\[ PV = nRT \]
Where:
- \( P = 1 \text{ atm} \)
- \( V \) is the volume we want to find
- \( n = 10 \text{ moles} \)
- \( R = 0.0821 \text{ L atm/(K mol)} \)
- \( T = 300 \text{ K} \)

Rearranging the ideal gas equation to solve for \( V \):
\[ V = \frac{nRT}{P} \]
Substituting the values:
\[ V = \frac{10 \text{ moles} \times 0.0821 \text{ L atm/(K mol)} \times 300 \text{ K}}{1 \text{ atm}} = 246.3 \text{ L} \]

### Step 5: Determine how many hours of service can be derived from the acetylene gas.
Given that the lamp requires 50 L of acetylene gas for 1 hour:
- For 1 hour, the lamp uses 50 L.
- We have 246.3 L of acetylene gas.

Using a unitary method:
\[ \text{Hours of service} = \frac{\text{Total volume of acetylene}}{\text{Volume required per hour}} = \frac{246.3 \text{ L}}{50 \text{ L/hour}} = 4.926 \text{ hours} \]

### Step 6: Round to the nearest whole number.
Thus, the number of hours of service that can be derived is approximately 5 hours.

### Final Answer:
**5 hours**

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