An electric discharge is passed through a mixture containing 50 c.c. of `O_(2)` and 50 c.c. of` H_(2)`. The volume of the gases formed (i) at room temperature and (ii) at `110^(@)`C will be

Options

A. (a)(i) 25 c.c. (ii) 50 c.c.
B. (b)(i) 50 c.c. (ii) 75 c.c.
C. (c )(i) 25 c.c. (ii) 75 c.c.
D. (d)(i) 75 c.c. (ii) 75 c.c.

Explanation

To solve the problem, we need to analyze the reaction between hydrogen (H₂) and oxygen (O₂) gases when an electric discharge is passed through their mixture. The reaction is as follows:

\[ \text{2H}_2 + \text{O}_2 \rightarrow \text{2H}_2\text{O} \]

### Step 1: Identify the initial volumes of the gases
We have:
- Volume of O₂ = 50 cm³
- Volume of H₂ = 50 cm³

### Step 2: Determine the limiting reagent
From the balanced equation, we see that:
- 2 volumes of H₂ react with 1 volume of O₂.

This means:
- 50 cm³ of H₂ would require \( \frac{50}{2} = 25 \) cm³ of O₂ to react completely.

Since we have 50 cm³ of O₂ available, H₂ is the limiting reagent because it will be completely consumed in the reaction.

### Step 3: Calculate the remaining volume of O₂ at room temperature
After the reaction:
- H₂ will be completely consumed (50 cm³).
- O₂ will be reduced by 25 cm³ (the amount that reacted).

Remaining volume of O₂:
\[ 50 \text{ cm³} - 25 \text{ cm³} = 25 \text{ cm³} \]

At room temperature (25°C), water (H₂O) produced will be in liquid form, so it does not contribute to the gas volume.

**Total gas volume at room temperature:**
- O₂ remaining = 25 cm³
- H₂ = 0 cm³ (consumed)
- Total = 25 cm³

### Step 4: Calculate the volume of gases at 110°C
At 110°C, water will be in gaseous form (steam). The reaction remains the same:
- H₂ is still the limiting reagent and will be completely consumed.
- The volume of O₂ remaining is still 25 cm³.

Now, the water produced (H₂O) will also be in gaseous form:
- From the balanced equation, 2 volumes of H₂ produce 2 volumes of H₂O.
- Therefore, 50 cm³ of H₂ will produce 50 cm³ of H₂O.

### Step 5: Calculate the total gas volume at 110°C
At 110°C:
- Remaining O₂ = 25 cm³
- H₂O (as steam) produced = 50 cm³

**Total gas volume at 110°C:**
\[ 25 \text{ cm³ (O₂)} + 50 \text{ cm³ (H₂O)} = 75 \text{ cm³} \]

### Final Answer:
- Volume of gases at room temperature = 25 cm³
- Volume of gases at 110°C = 75 cm³