How much charge (in F) must flow through solution during electrolysis of aq. `Na_(2)SO_(4)` at `STP` to produce `33.6 L` of product gases at `50%` current efficiency:

Explanation

To solve the problem of how much charge (in Faraday) must flow through a solution during the electrolysis of aqueous Na₂SO₄ to produce 33.6 L of product gases at 50% current efficiency, we can follow these steps:

### Step 1: Understand the Reaction
In the electrolysis of aqueous Na₂SO₄, water is oxidized at the anode to produce oxygen gas and hydrogen ions. The half-reaction at the anode can be represented as:
\[ 2H_2O \rightarrow 4H^+ + 4e^- + O_2 \]

### Step 2: Calculate Moles of Gas Produced
At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 L. Therefore, to find the number of moles of gas produced from 33.6 L, we can use the formula:
\[ \text{Moles of gas} = \frac{\text{Volume of gas}}{22.4 \, \text{L/mol}} \]
\[ \text{Moles of gas} = \frac{33.6 \, \text{L}}{22.4 \, \text{L/mol}} = 1.5 \, \text{mol} \]

### Step 3: Relate Moles of Gas to Faraday
From the half-reaction, we see that 1 mole of O₂ produced corresponds to 4 moles of electrons (4 Faraday). Therefore, for 1.5 moles of O₂, the total charge in Faraday can be calculated as:
\[ \text{Charge (F)} = \text{Moles of gas} \times 4 \, \text{F/mol} \]
\[ \text{Charge (F)} = 1.5 \, \text{mol} \times 4 \, \text{F/mol} = 6 \, \text{F} \]

### Step 4: Adjust for Current Efficiency
Since the current efficiency is 50%, we need to adjust the calculated charge:
\[ \text{Effective Charge (F)} = \text{Charge (F)} \times \frac{\text{Current Efficiency}}{100} \]
\[ \text{Effective Charge (F)} = 6 \, \text{F} \times \frac{50}{100} = 3 \, \text{F} \]

### Final Answer
The total charge that must flow through the solution during the electrolysis of aqueous Na₂SO₄ to produce 33.6 L of product gases at 50% current efficiency is **3 Faraday (F)**.

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