The magnitude of work done (in kcal) when 120 gm of Mg (s) is reacted with excess of hydrochloric acid in an open vessel at `27^@C` is

Explanation

To solve the problem of calculating the work done when 120 grams of magnesium reacts with excess hydrochloric acid in an open vessel at 27°C, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Reaction**:
The balanced chemical equation for the reaction between magnesium (Mg) and hydrochloric acid (HCl) is:
\[ Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g) \]

2. **Determine the Moles of Magnesium**:
The molar mass of magnesium (Mg) is approximately 24 g/mol. To find the number of moles of magnesium in 120 grams, use the formula:
\[ \text{Moles of Mg} = \frac{\text{mass}}{\text{molar mass}} = \frac{120 \text{ g}}{24 \text{ g/mol}} = 5 \text{ moles} \]

3. **Calculate Change in Moles of Gas (\( \Delta N_g \))**:
In the reaction, we have:
- Reactants: 2 moles of HCl (which is not a gas) + 1 mole of Mg (solid, not counted in gas)
- Products: 1 mole of \( H_2 \) (gas)

Therefore, the change in the number of moles of gas (\( \Delta N_g \)) is:
\[ \Delta N_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 1 - 0 = 1 \]

4. **Calculate Work Done (\( W \))**:
The work done in a reaction at constant pressure can be calculated using the formula:
\[ W = -P_{\text{external}} \Delta V \]
We can relate this to the change in moles of gas:
\[ W = -\Delta N_g \cdot R \cdot T \]
Where:
- \( R \) (gas constant) = 2 cal/(mol·K)
- \( T \) = 27°C = 300 K (since 27 + 273 = 300)

Substituting the values:
\[ W = -1 \cdot 2 \text{ cal/(mol·K)} \cdot 300 \text{ K} = -600 \text{ cal} \]

5. **Convert Work Done to Kilocalories**:
Since 1 kcal = 1000 cal, we convert the work done:
\[ W = -600 \text{ cal} \times \frac{1 \text{ kcal}}{1000 \text{ cal}} = -0.6 \text{ kcal} \]

### Final Answer:
The magnitude of work done when 120 grams of magnesium reacts with excess hydrochloric acid is:
\[ \boxed{-0.6 \text{ kcal}} \]