16.8 litre gas containing `H_(2)` and `O_(2)` is formed at NTP on electrolysis of water . What should be the weight of electrolysed water

Options

A. 5 g
B. 9 g
C. 10 g
D. 12 g

Explanation

To find the weight of the electrolyzed water that produces 16.8 liters of gas containing \( H_2 \) and \( O_2 \) at NTP (Normal Temperature and Pressure), we can follow these steps:

### Step 1: Determine the number of moles of gas produced
At NTP, 1 mole of any gas occupies 22.4 liters. Therefore, we can calculate the number of moles of gas produced using the formula:

\[ \text{Number of moles} = \frac{\text{Volume of gas}}{\text{Molar volume at NTP}} = \frac{16.8 \, \text{liters}}{22.4 \, \text{liters/mole}} \]

Calculating this gives:

\[ \text{Number of moles} = \frac{16.8}{22.4} = 0.75 \, \text{moles} \]

### Step 2: Determine the composition of the gas
During the electrolysis of water, the reaction produces hydrogen and oxygen in the ratio of 2:1. Thus, for every 2 moles of \( H_2 \), 1 mole of \( O_2 \) is produced.

Let’s denote:
- \( n_{H_2} \) = moles of \( H_2 \)
- \( n_{O_2} \) = moles of \( O_2 \)

From the stoichiometry of the reaction:

\[ n_{H_2} = 2n_{O_2} \]

Let \( n_{O_2} = x \). Then \( n_{H_2} = 2x \).

The total moles of gas produced is:

\[ n_{H_2} + n_{O_2} = 2x + x = 3x \]

Setting this equal to the total moles we found:

\[ 3x = 0.75 \implies x = 0.25 \]

Thus, we have:

\[ n_{O_2} = 0.25 \, \text{moles} \quad \text{and} \quad n_{H_2} = 0.5 \, \text{moles} \]

### Step 3: Calculate the total number of moles of water electrolyzed
The balanced equation for the electrolysis of water is:

\[ 2H_2O \rightarrow 2H_2 + O_2 \]

From the equation, we see that 2 moles of water produce 2 moles of \( H_2 \) and 1 mole of \( O_2 \). Therefore, the moles of water electrolyzed can be calculated as follows:

\[ \text{Moles of water} = \frac{2}{2} \times n_{H_2} = n_{H_2} = 0.5 \, \text{moles} \]

### Step 4: Calculate the weight of the electrolyzed water
The molar mass of water \( H_2O \) is approximately 18 g/mol. Therefore, the weight of the electrolyzed water can be calculated using:

\[ \text{Weight} = \text{Number of moles} \times \text{Molar mass} \]

Substituting the values:

\[ \text{Weight} = 0.5 \, \text{moles} \times 18 \, \text{g/mol} = 9 \, \text{grams} \]

### Final Answer
The weight of the electrolyzed water is **9 grams**.

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