The volume of `H_(2)` gas at NTP obtained by passing 4 amperes
through acidified `H_(2)O` for 30 minutes is

Options

A. 0.836L
B. 0.0432L
C. 0.1672L
D. 5.6L

Explanation

To solve the problem of finding the volume of \( H_2 \) gas at NTP obtained by passing 4 amperes through acidified \( H_2O \) for 30 minutes, we can follow these steps:

### Step 1: Calculate the total charge (Q) passed
The formula for charge is given by:
\[ Q = I \times t \]
where:
- \( I \) is the current in amperes (A)
- \( t \) is the time in seconds (s)

Given:
- \( I = 4 \, A \)
- \( t = 30 \, \text{minutes} = 30 \times 60 \, \text{seconds} = 1800 \, s \)

Now, substituting the values:
\[ Q = 4 \, A \times 1800 \, s = 7200 \, C \]

### Step 2: Relate charge to moles of \( H_2 \)
From electrolysis, we know that:
- 2 moles of electrons produce 1 mole of \( H_2 \).
- 1 mole of electrons corresponds to 1 Faraday of charge, which is approximately \( 96500 \, C \).

Thus, for 2 moles of electrons:
\[ \text{Charge for 1 mole of } H_2 = 2 \times 96500 \, C = 193000 \, C \]

### Step 3: Calculate the moles of \( H_2 \) produced
Using the total charge calculated:
\[ \text{Moles of } H_2 = \frac{Q}{\text{Charge for 1 mole of } H_2} = \frac{7200 \, C}{193000 \, C} \approx 0.0373 \, \text{moles} \]

### Step 4: Calculate the volume of \( H_2 \) at NTP
At Normal Temperature and Pressure (NTP), 1 mole of any ideal gas occupies 22.4 liters. Therefore, the volume of \( H_2 \) produced can be calculated as:
\[ \text{Volume of } H_2 = \text{Moles of } H_2 \times 22.4 \, \text{L/mol} = 0.0373 \, \text{moles} \times 22.4 \, \text{L/mol} \approx 0.836 \, \text{L} \]

### Final Answer
The volume of \( H_2 \) gas at NTP obtained is approximately \( 0.836 \, \text{liters} \).

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