The work done in an open vessel at `300K`, when `112g` iron reacts with dil. `HCL` is:

Options

A. `1200 cal`
B. `600 cal`
C. `300 cal`
D. `200 cal`

Explanation

To find the work done in an open vessel at 300 K when 112 g of iron reacts with dilute HCl, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction of iron (Fe) with dilute hydrochloric acid (HCl) can be represented as:
\[ \text{Fe (s)} + 2 \text{HCl (aq)} \rightarrow \text{FeCl}_2 (aq) + \text{H}_2 (g) \]

### Step 2: Calculate the number of moles of iron
To find the number of moles of iron, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
Given that the mass of iron is 112 g and the molar mass of iron (Fe) is 56 g/mol:
\[ \text{Number of moles of Fe} = \frac{112 \, \text{g}}{56 \, \text{g/mol}} = 2 \, \text{moles} \]

### Step 3: Determine the moles of hydrogen gas produced
From the balanced equation, we see that 1 mole of Fe produces 1 mole of H2 gas. Therefore, 2 moles of Fe will produce:
\[ 2 \, \text{moles of H}_2 \]

### Step 4: Calculate the change in volume
In an open vessel, the initial volume due to solid (Fe) and liquid (HCl) is negligible compared to the volume of gas produced. Thus, the change in volume (ΔV) is equal to the volume of the hydrogen gas produced.

### Step 5: Use the ideal gas law to find the volume of hydrogen gas
Using the ideal gas equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure (assumed to be 1 atm)
- \( V \) = volume of gas
- \( n \) = number of moles of gas (2 moles of H2)
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin (300 K)

Rearranging to find volume:
\[ V = \frac{nRT}{P} \]
Substituting the values:
\[ V = \frac{2 \, \text{moles} \times 0.0821 \, \text{L·atm/(K·mol)} \times 300 \, \text{K}}{1 \, \text{atm}} \]
\[ V = \frac{49.26}{1} = 49.26 \, \text{L} \]

### Step 6: Calculate the work done
The work done (W) in an open vessel can be calculated using:
\[ W = P \Delta V \]
Since we already have \( V \) and \( P \) is 1 atm:
\[ W = 1 \, \text{atm} \times 49.26 \, \text{L} \]
To convert this to calories, we use the conversion factor \( 1 \, \text{L·atm} = 24.2 \, \text{calories} \):
\[ W = 49.26 \, \text{L} \times 24.2 \, \text{calories/L·atm} \]
\[ W = 1195.59 \, \text{calories} \approx 1200 \, \text{calories} \]

### Final Answer
The work done in an open vessel at 300 K when 112 g of iron reacts with dilute HCl is approximately **1200 calories**.

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