If 270.0 g of water is electrolysed during an experiment performed by miss abhilasha with 75% current efficiency then

Options

A. 168 L of `O_2` (g) will be evolved at anode at 1 atm & 273 K
B. Total 504 L gases will be produced at 1 atm & 273 K
C. 336 L of `H_2` (g) will be evolved at anode at 1 atm & 273 K
D. 45 F electricity will be consumed

Answer

To solve the problem of electrolysis of water, we can follow these steps:

### Step 1: Calculate the moles of water
Given the mass of water (H₂O) is 270.0 g, we can find the number of moles using the formula:

\[ \text{Moles of H₂O} = \frac{\text{mass}}{\text{molar mass}} \]

The molar mass of water (H₂O) is approximately 18 g/mol.

\[ \text{Moles of H₂O} = \frac{270.0 \, \text{g}}{18 \, \text{g/mol}} = 15 \, \text{moles} \]

**Hint:** Remember to use the molar mass of water to convert grams to moles.

### Step 2: Determine the equivalents of water
The electrolysis of water produces hydrogen and oxygen in a 2:1 ratio. Therefore, the number of equivalents of water can be calculated as follows:

\[ \text{Equivalents of H₂O} = \text{Moles of H₂O} \times 2 = 15 \times 2 = 30 \, \text{equivalents} \]

**Hint:** The stoichiometric coefficients in the balanced equation for the electrolysis of water are crucial for determining equivalents.

### Step 3: Calculate the volume of oxygen evolved
From the balanced reaction, for every 2 moles of water, 1 mole of oxygen is produced. Therefore, the moles of oxygen produced will be:

\[ \text{Moles of O₂} = \frac{\text{Moles of H₂O}}{2} = \frac{15}{2} = 7.5 \, \text{moles} \]

Now, we can calculate the volume of oxygen at STP (Standard Temperature and Pressure, where 1 mole of gas occupies 22.4 L):

\[ \text{Volume of O₂} = \text{Moles of O₂} \times 22.4 \, \text{L/mol} = 7.5 \times 22.4 = 168 \, \text{L} \]

**Hint:** Use the ideal gas law at STP to find the volume of gases produced.

### Step 4: Calculate the total volume of gases produced
The total volume of gases produced includes both hydrogen and oxygen. For every 2 moles of water, 2 moles of hydrogen and 1 mole of oxygen are produced, giving us:

\[ \text{Total moles of gases} = \text{Moles of H₂} + \text{Moles of O₂} = 15 \times 2 + 7.5 = 30 + 7.5 = 37.5 \, \text{moles} \]

Now, we can calculate the total volume of gases at STP:

\[ \text{Total volume of gases} = \text{Total moles of gases} \times 22.4 \, \text{L/mol} = 37.5 \times 22.4 = 840 \, \text{L} \]

**Hint:** Remember to account for both hydrogen and oxygen when calculating the total volume of gases produced.

### Step 5: Calculate the Faraday's electricity consumed
The total charge (in Faradays) required to produce the equivalents of hydrogen and oxygen can be calculated using the current efficiency. The theoretical charge required is given by:

\[ \text{Faraday's required} = \text{Equivalents} \times \text{Current Efficiency} \]

Given the current efficiency is 75%, we can calculate the actual charge consumed:

\[ \text{Faraday's consumed} = \frac{30}{0.75} = 40 \, \text{Faradays} \]

**Hint:** Current efficiency affects the actual charge consumed, so adjust your calculations accordingly.

### Conclusion
Based on the calculations:
1. 168 liters of oxygen will be evolved at anode at STP.
2. The total volume of gases produced is 840 liters.
3. The total Faraday's electricity consumed is 40 Faradays.