`4 g H_(2)` &`127 g I_(2)` are mixed & heated in 10 lit closed vessel until equilibrium is reached. If the equilibrium concentration of HI is 0.05 M, total number of moles present at cquilibrium is

Options

A. a) 3.25
B. b) 1.75
C. c) 2.25
D. d) 2.5

Answer

To solve the problem, we need to determine the total number of moles present at equilibrium after mixing 4 g of H₂ and 127 g of I₂ in a 10-liter closed vessel, given that the equilibrium concentration of HI is 0.05 M.

### Step-by-Step Solution:

1. **Calculate the number of moles of H₂ and I₂:**
- The molar mass of H₂ is 2 g/mol.
- The number of moles of H₂:
\[ \text{Moles of H₂} = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \text{ g}}{2 \text{ g/mol}} = 2 \text{ moles} \]

- The molar mass of I₂ is 254 g/mol.
- The number of moles of I₂:
\[ \text{Moles of I₂} = \frac{127 \text{ g}}{254 \text{ g/mol}} = 0.5 \text{ moles} \]

2. **Write the balanced chemical equation:**
\[ \text{H₂} + \text{I₂} \rightleftharpoons 2 \text{HI} \]

3. **Set up the initial conditions:**
- Initial moles:
- H₂ = 2 moles
- I₂ = 0.5 moles
- HI = 0 moles

4. **Define the change in moles at equilibrium:**
- Let \( \alpha \) be the change in moles of H₂ and I₂ that react.
- At equilibrium:
- Moles of H₂ = \( 2 - \alpha \)
- Moles of I₂ = \( 0.5 - \alpha \)
- Moles of HI = \( 2\alpha \)

5. **Use the given equilibrium concentration of HI:**
- The equilibrium concentration of HI is given as 0.05 M in a 10-liter vessel.
- Calculate the moles of HI at equilibrium:
\[ \text{Moles of HI} = \text{Concentration} \times \text{Volume} = 0.05 \text{ M} \times 10 \text{ L} = 0.5 \text{ moles} \]

6. **Relate the moles of HI to \( \alpha \):**
- From the equilibrium expression for HI:
\[ 2\alpha = 0.5 \]
- Therefore, solving for \( \alpha \):
\[ \alpha = \frac{0.5}{2} = 0.25 \]

7. **Calculate the moles of H₂ and I₂ at equilibrium:**
- Moles of H₂ at equilibrium:
\[ \text{Moles of H₂} = 2 - \alpha = 2 - 0.25 = 1.75 \text{ moles} \]
- Moles of I₂ at equilibrium:
\[ \text{Moles of I₂} = 0.5 - \alpha = 0.5 - 0.25 = 0.25 \text{ moles} \]

8. **Total moles at equilibrium:**
- Total moles = Moles of H₂ + Moles of I₂ + Moles of HI
\[ \text{Total moles} = 1.75 + 0.25 + 0.5 = 2.5 \text{ moles} \]

### Final Answer:
The total number of moles present at equilibrium is **2.5 moles**.