When `22.4 L` of `H_(2)(g)` is mixed with 11.2 of `Cl_(2)(g)`, each at STP, the moles of HCl(g) formed is equal to

Options

A. (a)1 mole of `HCl(g)`
B. (b)2 moles of `HCl(g)`
C. (c )0.5 mole of `HCl(g)`
D. (d)1.5 mole of `HCl(g)`

Explanation

To determine the moles of HCl(g) formed when 22.4 L of H₂(g) is mixed with 11.2 L of Cl₂(g) at STP, we follow these steps:

1. **Understand the reaction and stoichiometry:**
The balanced chemical equation for the reaction between hydrogen gas (H₂) and chlorine gas (Cl₂) to form hydrogen chloride gas (HCl) is:
\[ \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g) \]

2. **Volume to moles conversion at STP:**
At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore:
- 22.4 L of H₂ corresponds to 1 mole of H₂.
- 11.2 L of Cl₂ corresponds to 0.5 moles of Cl₂.

3. **Identify the limiting reagent:**
According to the stoichiometry of the reaction, 1 mole of H₂ reacts with 1 mole of Cl₂ to produce 2 moles of HCl. However, we have:
- 1 mole of H₂
- 0.5 moles of Cl₂

Since the reaction requires 1 mole of Cl₂ to completely react with 1 mole of H₂, and we only have 0.5 moles of Cl₂, Cl₂ is the limiting reagent.

4. **Calculate the moles of HCl produced:**
The limiting reagent determines the amount of product formed. According to the balanced equation, 1 mole of Cl₂ produces 2 moles of HCl. Therefore, 0.5 moles of Cl₂ will produce:
\[ 0.5 \text{ moles of Cl}_2 \times \frac{2 \text{ moles of HCl}}{1 \text{ mole of Cl}_2} = 1 \text{ mole of HCl} \]

Thus, the moles of HCl(g) formed is 1 mole.