In the electrolysis of aqueous `NaCl` ,what volume of `Cl_2`(g) is produced in the time that it takes to liberate 5.0 liter of `H_2`(g) ? Assume that both gases are measured at STP.

Options

A. 5
B. 2.5
C. 7.5
D. 10

Answer

To solve the problem, we need to analyze the electrolysis of aqueous sodium chloride (NaCl) and the stoichiometry of the gases produced during the process.

### Step-by-Step Solution:

1. **Understanding the Electrolysis Reaction**:
During the electrolysis of aqueous NaCl, the following reactions occur:
- At the anode (oxidation): Chloride ions (Cl⁻) are oxidized to chlorine gas (Cl₂).
- At the cathode (reduction): Water is reduced to hydrogen gas (H₂).

The overall balanced reaction can be represented as:
\[ 2 \text{NaCl} + 2 \text{H}_2\text{O} \rightarrow \text{Cl}_2 + \text{H}_2 + 2 \text{NaOH} \]

2. **Identifying the Molar Ratios**:
From the balanced equation, we see that:
- 1 mole of Cl₂ is produced for every 1 mole of H₂ produced.

3. **Volume of Gases at STP**:
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of gases produced is directly proportional to the number of moles.

4. **Calculating the Volume of Cl₂ Produced**:
Given that 5.0 liters of H₂ is produced, we can determine the volume of Cl₂ produced using the 1:1 molar ratio:
- Since 5.0 liters of H₂ corresponds to 5.0 liters of Cl₂ (because of the 1:1 ratio from the balanced equation).

5. **Final Answer**:
Therefore, the volume of Cl₂ produced is:
\[ \text{Volume of Cl}_2 = 5.0 \text{ liters} \]

### Summary:
In the electrolysis of aqueous NaCl, when 5.0 liters of H₂ is produced, an equal volume of Cl₂ (5.0 liters) is also produced.