0.01 mole of iodoform `(CHI_(3))` reacts with Ag to produce a gas whose volume at NTP is
`2CHI_(3)+6Ag to6Agl(s)+C_(2)H_(2)(g)`

Options

A. 224 mL
B. 112 mL
C. 336 mL
D. None of these

Answer

To solve the problem, we need to determine the volume of acetylene gas (C₂H₂) produced when 0.01 moles of iodoform (CHI₃) reacts with silver (Ag) according to the given reaction:

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

**Step 1: Determine the moles of C₂H₂ produced from the moles of CHI₃.**

From the balanced chemical equation, we see that:
- 2 moles of CHI₃ produce 1 mole of C₂H₂.

Thus, the relationship can be expressed as:
\[ \text{Moles of C}_2\text{H}_2 = \frac{1}{2} \times \text{Moles of CHI}_3 \]

Given that we have 0.01 moles of CHI₃:
\[ \text{Moles of C}_2\text{H}_2 = \frac{1}{2} \times 0.01 = 0.005 \text{ moles} \]

**Step 2: Calculate the volume of C₂H₂ at NTP.**

At Normal Temperature and Pressure (NTP), 1 mole of any gas occupies 22.4 liters. Therefore, we can calculate the volume of 0.005 moles of C₂H₂ as follows:
\[ \text{Volume of C}_2\text{H}_2 = \text{Moles of C}_2\text{H}_2 \times 22.4 \text{ L/mol} \]
\[ \text{Volume of C}_2\text{H}_2 = 0.005 \times 22.4 = 0.112 \text{ liters} \]

**Step 3: Convert the volume from liters to milliliters.**

Since 1 liter is equal to 1000 milliliters:
\[ 0.112 \text{ liters} = 0.112 \times 1000 = 112 \text{ mL} \]

Thus, the final answer is:
**The volume of C₂H₂ produced at NTP is 112 mL.**

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