In a reaction vessel, `100 g H_(2)` and 100 g `Cl_(2)` are mixed and suitable conditions are provided for the reaction: `H_(2(g))+Cl_(2(g))to2HCl_((g))`
The amount of HCl formed (at 90% yield) will be

Options

A. 36.8 g
B. 62.5 g
C. 80g
D. 92.53 g

Answer

To solve the problem of how much HCl is formed when 100 g of H₂ and 100 g of Cl₂ are mixed under suitable conditions, we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2 \text{HCl}(g) \]

### Step 2: Calculate the number of moles of H₂ and Cl₂
- **For H₂:**
- Molar mass of H₂ = 2 g/mol
- Given mass = 100 g
- Number of moles of H₂ = \(\frac{\text{mass}}{\text{molar mass}} = \frac{100 \text{ g}}{2 \text{ g/mol}} = 50 \text{ moles}\)

- **For Cl₂:**
- Molar mass of Cl₂ = 2 × 35.5 g/mol = 71 g/mol
- Given mass = 100 g
- Number of moles of Cl₂ = \(\frac{100 \text{ g}}{71 \text{ g/mol}} \approx 1.41 \text{ moles}\)

### Step 3: Identify the limiting reagent
- From the stoichiometry of the reaction, 1 mole of Cl₂ reacts with 1 mole of H₂ to produce 2 moles of HCl.
- Since we have 1.41 moles of Cl₂ and 50 moles of H₂, Cl₂ is the limiting reagent because it has fewer moles.

### Step 4: Calculate the theoretical yield of HCl
- According to the reaction, 1 mole of Cl₂ produces 2 moles of HCl.
- Therefore, 1.41 moles of Cl₂ will produce:
\[ 2 \times 1.41 \text{ moles of HCl} = 2.82 \text{ moles of HCl} \]

### Step 5: Adjust for the yield
- The problem states that the yield is 90%. Therefore, the actual moles of HCl produced will be:
\[ 2.82 \text{ moles} \times 0.90 = 2.538 \text{ moles} \]

### Step 6: Calculate the mass of HCl produced
- Molar mass of HCl = 1 (H) + 35.5 (Cl) = 36.5 g/mol
- The mass of HCl produced is:
\[ \text{mass} = \text{number of moles} \times \text{molar mass} = 2.538 \text{ moles} \times 36.5 \text{ g/mol} \approx 92.5 \text{ g} \]

### Final Answer
The amount of HCl formed at 90% yield will be approximately **92.5 g**.

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