`H_2`(g) and `O_2`(g), can be produced by the electrolysis of water. What total volume (in L) of `O_2` and `H_2` are produced at 1 atm and 273K when a current of 30 A is passed through a `K_2SO_4` (aq) solution for 193 min?

Options

A. 120.96
B. 40.32
C. 60.48
D. 80.64

Answer

To solve the problem, we need to calculate the total volume of hydrogen (`H_2`) and oxygen (`O_2`) produced during the electrolysis of water when a current of 30 A is passed through a `K_2SO_4` solution for 193 minutes.

### Step-by-Step Solution:

1. **Convert Time from Minutes to Seconds**:
\[ \text{Time (s)} = 193 \text{ minutes} \times 60 \text{ seconds/minute} = 11580 \text{ seconds} \]

**Hint**: Remember that there are 60 seconds in a minute.

2. **Calculate the Total Charge (Q)**:
Using the formula \( Q = I \times t \):
\[ Q = 30 \text{ A} \times 11580 \text{ s} = 347400 \text{ C} \]

**Hint**: Current (A) multiplied by time (s) gives you charge in coulombs (C).

3. **Determine the Number of Faradays (F)**:
One Faraday corresponds to 96500 C. To find the number of Faradays used:
\[ \text{Number of Faradays} = \frac{Q}{96500} = \frac{347400}{96500} \approx 3.6 \text{ Faradays} \]

**Hint**: 1 Faraday is the charge required to produce 1 mole of hydrogen.

4. **Calculate Moles of Hydrogen Produced**:
From the electrolysis of water, 1 mole of `H_2` is produced per 2 Faradays. Therefore:
\[ \text{Moles of } H_2 = \frac{3.6 \text{ Faradays}}{2} = 1.8 \text{ moles} \]

**Hint**: Remember the stoichiometry of the reaction: 2 Faradays produce 1 mole of hydrogen.

5. **Calculate Volume of Hydrogen at STP**:
At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 L. Thus:
\[ \text{Volume of } H_2 = 1.8 \text{ moles} \times 22.4 \text{ L/mole} = 40.32 \text{ L} \]

**Hint**: Use the molar volume of a gas at STP (22.4 L) to find the volume produced.

6. **Calculate Moles of Oxygen Produced**:
From the electrolysis of water, 1 mole of `O_2` is produced per 4 Faradays. Therefore:
\[ \text{Moles of } O_2 = \frac{3.6 \text{ Faradays}}{4} = 0.9 \text{ moles} \]

**Hint**: Recall that 4 Faradays are needed to produce 1 mole of oxygen.

7. **Calculate Volume of Oxygen at STP**:
\[ \text{Volume of } O_2 = 0.9 \text{ moles} \times 22.4 \text{ L/mole} = 20.16 \text{ L} \]

**Hint**: Again, apply the molar volume of a gas at STP.

8. **Calculate Total Volume of `H_2` and `O_2`**:
\[ \text{Total Volume} = \text{Volume of } H_2 + \text{Volume of } O_2 = 40.32 \text{ L} + 20.16 \text{ L} = 60.48 \text{ L} \]

**Hint**: Simply add the volumes of hydrogen and oxygen to find the total.

### Final Answer:
The total volume of `H_2` and `O_2` produced is **60.48 liters**.