3.94 g of iodoform is heated with Ag powder. What will be the volume of the gas obtained at NTP?

Options

A. 224 ml.
B. 112 ml
C. 22400 ml.
D. 2240 ml

Answer

To solve the problem of finding the volume of gas obtained when 3.94 g of iodoform (CHI3) is heated with silver powder (Ag), we can follow these steps:

### Step 1: Determine the molecular weight of iodoform (CHI3)
The molecular weight of iodoform can be calculated as follows:
- Carbon (C) = 12 g/mol
- Hydrogen (H) = 1 g/mol
- Iodine (I) = 127 g/mol

Molecular weight of CHI3 = 12 + 1 + (3 × 127)
= 12 + 1 + 381
= 394 g/mol

### Step 2: Calculate the number of moles of iodoform (CHI3)
Using the formula for moles:
\[ \text{Moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

Substituting the values:
\[ \text{Moles of CHI3} = \frac{3.94 \text{ g}}{394 \text{ g/mol}} = 0.01 \text{ moles} \]

### Step 3: Write the balanced chemical equation
The reaction between iodoform and silver powder can be represented as:
\[ 2 \text{CHI}_3 + 6 \text{Ag} \rightarrow 6 \text{AgI} + \text{C}_2\text{H}_2 \]

From the equation, we see that:
- 2 moles of CHI3 produce 1 mole of C2H2 (acetylene).

### Step 4: Calculate the moles of acetylene (C2H2) produced
From the stoichiometry of the reaction:
\[ \text{Moles of C2H2} = \frac{0.01 \text{ moles of CHI3}}{2} = 0.005 \text{ moles of C2H2} \]

### Step 5: Calculate the volume of gas produced at NTP
At Normal Temperature and Pressure (NTP), 1 mole of gas occupies 22.4 liters (or 22400 mL).

To find the volume of 0.005 moles of C2H2:
\[ \text{Volume} = \text{moles} \times \text{volume per mole} \]
\[ \text{Volume} = 0.005 \text{ moles} \times 22400 \text{ mL/mole} \]
\[ \text{Volume} = 112 \text{ mL} \]

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