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 of calculating the number of hours of service that can be derived from an acetylene lamp containing 640 g of calcium carbide, we can follow these steps:

### Step 1: Write the Reaction
The reaction for the decomposition of calcium carbide (CaC₂) to produce acetylene (C₂H₂) is as follows:
\[ \text{CaC}_2 + \text{H}_2O \rightarrow \text{C}_2H_2 + \text{Ca(OH)}_2 \]
This shows that one mole of calcium carbide produces one mole of acetylene.

### Step 2: Calculate the Molar Mass of Calcium Carbide
To find the number of moles of calcium carbide, we first need to calculate its molar mass.
- Calcium (Ca) = 40 g/mol
- Carbon (C) = 12 g/mol (and there are 2 carbon atoms in calcium carbide)

Thus, the molar mass of calcium carbide (CaC₂) is:
\[ \text{Molar mass of CaC}_2 = 40 + (2 \times 12) = 40 + 24 = 64 \text{ g/mol} \]

### Step 3: Calculate the Number of Moles of Calcium Carbide
Using the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
Given mass of calcium carbide = 640 g, we can calculate:
\[ \text{Number of moles} = \frac{640 \text{ g}}{64 \text{ g/mol}} = 10 \text{ moles} \]

### Step 4: Calculate the Volume of Acetylene Produced
Using the ideal gas equation \( PV = nRT \), we can rearrange it to find the volume (V):
\[ V = \frac{nRT}{P} \]
Where:
- \( n = 10 \text{ moles} \)
- \( R = 0.0821 \text{ L atm/(K mol)} \)
- \( T = 300 \text{ K} \)
- \( P = 1 \text{ atm} \)

Substituting the values:
\[ V = \frac{10 \times 0.0821 \times 300}{1} \]
Calculating \( 0.0821 \times 300 = 24.63 \):
\[ V = 10 \times 24.63 = 246.3 \text{ L} \]

### Step 5: Determine the Hours of Service
The lamp requires 50 L of acetylene gas for one hour. To find the total hours of service, we can use a unitary method:
\[ \text{Hours of service} = \frac{\text{Total volume of acetylene}}{\text{Volume required per hour}} \]
\[ \text{Hours of service} = \frac{246.3 \text{ L}}{50 \text{ L/hour}} = 4.926 \text{ hours} \]

### Step 6: Round to the Nearest Whole Number
Rounding 4.926 to the nearest whole number gives us approximately 5 hours.

### Final Answer
The number of hours of service that can be derived from the acetylene lamp containing 640 g of calcium carbide is approximately **5 hours**.

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