The volume of ascetylene at `NTP` produced by reaction of `50 gm` of `CaC_(2)` with water is : `CaC_(2)+H_(2)O rarr Ca(OH)_(2)+C_(2)H_(2)`

Options

A. `10L`
B. `17.5L`
C. `26.25L`
D. `35L`

Explanation

To solve the question regarding the volume of acetylene (C₂H₂) produced from the reaction of 50 grams of calcium carbide (CaC₂) with water, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between calcium carbide and water is given as:
\[ \text{CaC}_2 + 2 \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{C}_2\text{H}_2 \]

### Step 2: Calculate the molar mass of CaC₂
The molar mass of calcium carbide (CaC₂) can be calculated as follows:
- Calcium (Ca) = 40 g/mol
- Carbon (C) = 12 g/mol (and there are 2 carbon atoms)

Thus, the molar mass of 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 CaC₂
Using the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]

Substituting the values:
\[ \text{Number of moles of CaC}_2 = \frac{50 \text{ g}}{64 \text{ g/mol}} = 0.78125 \text{ moles} \]

### Step 4: Determine the moles of C₂H₂ produced
From the balanced equation, we see that 1 mole of CaC₂ produces 1 mole of C₂H₂. Therefore, the number of moles of C₂H₂ produced will also be:
\[ \text{Moles of C}_2\text{H}_2 = 0.78125 \text{ moles} \]

### Step 5: Calculate the volume of C₂H₂ at NTP
At Normal Temperature and Pressure (NTP), 1 mole of gas occupies 22.4 liters. Thus, the volume of C₂H₂ produced can be calculated as:
\[ \text{Volume} = \text{Number of moles} \times \text{Molar volume} \]
\[ \text{Volume of C}_2\text{H}_2 = 0.78125 \text{ moles} \times 22.4 \text{ L/mol} \]

Calculating this gives:
\[ \text{Volume of C}_2\text{H}_2 = 17.5 \text{ liters} \]

### Final Answer
The volume of acetylene produced at NTP from the reaction of 50 grams of CaC₂ with water is **17.5 liters**.

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