To calculate the work done when 112 g of iron reacts with dilute HCl at 300 K, 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:
\[ \text{Fe} + 2 \text{HCl} \rightarrow \text{FeCl}_2 + \text{H}_2 \]
### Step 2: Determine the moles of iron
Given that the mass of iron (Fe) is 112 g and the molar mass of iron is 56 g/mol, we can calculate the number of moles of iron:
\[
\text{Moles of Fe} = \frac{\text{mass of Fe}}{\text{molar mass of Fe}} = \frac{112 \, \text{g}}{56 \, \text{g/mol}} = 2 \, \text{moles}
\]
### Step 3: Determine the moles of hydrogen produced
From the balanced equation, 1 mole of iron produces 1 mole of hydrogen. Therefore, 2 moles of iron will produce 2 moles of hydrogen:
\[
\text{Moles of } H_2 = 2 \, \text{moles}
\]
### Step 4: Calculate the volume of hydrogen produced at 300 K
Using the ideal gas law, \( PV = nRT \), we can find the volume of hydrogen produced. Rearranging gives:
\[
V = \frac{nRT}{P}
\]
Assuming the reaction occurs at standard atmospheric pressure (1 atm = 101.325 kPa), we can use:
- \( n = 2 \, \text{moles} \)
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 300 \, \text{K} \)
Substituting these values in:
\[
V = \frac{2 \, \text{moles} \times 8.314 \, \text{J/(mol K)} \times 300 \, \text{K}}{101.325 \, \text{kPa}}
\]
Converting kPa to J/L (1 kPa = 1000 J/m³):
\[
V = \frac{2 \times 8.314 \times 300}{101.325 \times 1000} \approx 0.04988 \, \text{m}^3 = 49.88 \, \text{L}
\]
### Step 5: Calculate the work done
The work done in a constant pressure process is given by:
\[
W = -P \Delta V
\]
Where \( \Delta V = V_f - V_i \). Since the initial volume \( V_i = 0 \) (no gas before the reaction), we have:
\[
W = -P V_f
\]
Substituting \( P = 101.325 \, \text{kPa} = 101325 \, \text{Pa} \) and \( V_f \approx 0.04988 \, \text{m}^3 \):
\[
W = -101325 \, \text{Pa} \times 0.04988 \, \text{m}^3 \approx -5048.6 \, \text{J}
\]
### Step 6: Convert work done to calories
To convert joules to calories, we use the conversion factor \( 1 \, \text{cal} = 4.184 \, \text{J} \):
\[
W \approx \frac{-5048.6 \, \text{J}}{4.184 \, \text{J/cal}} \approx -1200 \, \text{cal}
\]
### Final Answer
The work done when 112 g of iron reacts with dilute HCl at 300 K is approximately:
\[
\boxed{-1200 \, \text{cal}}
\]
