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 in this reaction (at 100% yield) will be

Options

A. 102.8 g
B. 73 g
C. 36.5 g
D. 142 g

Answer

To solve the problem, we need to determine the amount of HCl formed when 100 g of H₂ and 100 g of Cl₂ react completely. We will 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
To find the number of moles of H₂, we use the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]

The molar mass of H₂ is 2 g/mol. Therefore, the number of moles of H₂ is:
\[ \text{Number of moles of H}_2 = \frac{100 \text{ g}}{2 \text{ g/mol}} = 50 \text{ moles} \]

### Step 3: Calculate the number of moles of Cl₂
Similarly, we calculate the number of moles of Cl₂. The molar mass of Cl₂ (Chlorine) is approximately 71 g/mol (35.5 g/mol for Cl × 2). Thus, the number of moles of Cl₂ is:
\[ \text{Number of moles of Cl}_2 = \frac{100 \text{ g}}{71 \text{ g/mol}} \approx 1.41 \text{ moles} \]

### Step 4: Identify the limiting reagent
From the balanced equation, we see that 1 mole of Cl₂ reacts with 1 mole of H₂ to produce 2 moles of HCl. Since we have:
- 50 moles of H₂
- 1.41 moles of Cl₂

Cl₂ is the limiting reagent because it is present in a smaller amount compared to the stoichiometric requirement.

### Step 5: Calculate the number of moles of HCl produced
According to the balanced equation, 1 mole of Cl₂ produces 2 moles of HCl. Therefore, the number of moles of HCl produced from 1.41 moles of Cl₂ is:
\[ \text{Number of moles of HCl} = 2 \times \text{Number of moles of Cl}_2 = 2 \times 1.41 \approx 2.82 \text{ moles} \]

### Step 6: Calculate the mass of HCl produced
To find the mass of HCl produced, we first need the molar mass of HCl. The molar mass of HCl is:
\[ \text{Molar mass of HCl} = 1 \text{ g/mol (H)} + 35.5 \text{ g/mol (Cl)} = 36.5 \text{ g/mol} \]

Now, we can calculate the mass of HCl produced:
\[ \text{Mass of HCl} = \text{Number of moles of HCl} \times \text{Molar mass of HCl} \]
\[ \text{Mass of HCl} = 2.82 \text{ moles} \times 36.5 \text{ g/mol} \approx 102.93 \text{ g} \]

### Final Answer
The amount of HCl formed in this reaction (at 100% yield) will be approximately **102.93 g**.

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