Question

When 22.4 L of $H_{2(g)}$ is mixed with 11.2 L of $Cl_{2(g)}$, each at STP, the moles of $HCl_{(g)}$ formed is equal to

• A. 1 mole of $HCl_{(g)}$
• B. 2 moles of $HCl_{(g)}$
• C. 0.5 mole of $HCl_{(g)}$
• D. 1.5 moles of $HCl_{(g)}$

Answer 1

The Correct Option is A

To determine the number of moles of HCl_{(g)} formed when 22.4 L of H_2_{(g)} is mixed with 11.2 L of Cl_2_{(g)} at Standard Temperature and Pressure (STP), let's follow the steps below:

Step 1: Write the balanced chemical equation for the reaction.

The reaction between hydrogen gas and chlorine gas to form hydrogen chloride gas is represented by the balanced chemical equation:

H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}

Step 2: Use the concept of molar volume at STP.

At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of the gases can be directly related to moles.

Step 3: Calculate the moles of reactants.

• The moles of H_{2(g)}: \text{Moles of } H_2 = \frac{22.4 \, \text{L}}{22.4 \, \text{L/mol}} = 1 \, \text{mol}
• The moles of Cl_{2(g)}: \text{Moles of } Cl_2 = \frac{11.2 \, \text{L}}{22.4 \, \text{L/mol}} = 0.5 \, \text{mol}

Step 4: Determine the limiting reagent.

The balanced equation shows that 1 mole of H_2 reacts with 1 mole of Cl_2 to produce 2 moles of HCl. Given the moles:

• H_2 available = 1 mole
• Cl_2 available = 0.5 moles

Since Cl_2 is available in lesser moles compared to required stoichiometry, it acts as the limiting reagent.

Step 5: Calculate the moles of HCl produced.

The reaction produces 2 moles of HCl for every 1 mole of Cl_2:

\text{Moles of } HCl = 2 \times \text{moles of } Cl_2 = 2 \times 0.5 = 1 \, \text{mole}

Therefore, the moles of HCl_{(g)} formed is equal to 1 mole.

Conclusion:

The correct answer is 1 mole of HCl_{(g)}.