0.01 mole of idoform `(CHl_(3))` reacts with excess Ag power to produce a gas whose volume at NTP is

Options

A. 224 ml
B. 112 ml
C. 336 ml
D. 448 ml

Explanation

To solve the problem, we need to determine the volume of gas produced when 0.01 moles of iodophor (CHI₃) reacts with excess silver powder (Ag). Here’s the step-by-step solution:

### Step 1: Write the balanced chemical equation
The reaction of iodophor (CHI₃) with excess silver powder (Ag) can be represented as follows:

\[ 2 \text{CHI}_3 + 6 \text{Ag} \rightarrow 6 \text{AgI} + \text{C}_2\text{H}_2 \]

From this balanced equation, we see that 2 moles of CHI₃ produce 1 mole of acetylene gas (C₂H₂).

### Step 2: Determine the moles of C₂H₂ produced
Given that we have 0.01 moles of CHI₃, we can use stoichiometry to find out how many moles of C₂H₂ will be produced:

\[ \text{Moles of C}_2\text{H}_2 = \frac{1}{2} \times \text{Moles of CHI}_3 = \frac{1}{2} \times 0.01 = 0.005 \text{ moles} \]

### Step 3: Calculate the volume of C₂H₂ at NTP
At Normal Temperature and Pressure (NTP), 1 mole of gas occupies 22.4 liters. Therefore, the volume of 0.005 moles of C₂H₂ can be calculated as follows:

\[ \text{Volume} = \text{Moles} \times \text{Volume per mole} = 0.005 \text{ moles} \times 22.4 \text{ L/mole} = 0.112 \text{ L} \]

### Step 4: Convert volume from liters to milliliters
Since the question asks for the volume in milliliters, we need to convert liters to milliliters:

\[ \text{Volume in mL} = 0.112 \text{ L} \times 1000 \text{ mL/L} = 112 \text{ mL} \]

### Final Answer
The volume of gas produced at NTP is **112 mL**.

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