8 gm `H_(2) and 32 gm O_(2)` is allowed to recant to form water then which of the following statement is correct ?

Options

A. `O_(2)` is limiting reagent
B. `O_(2)` is reagent in excess
C. `H_(2)` is limiting reagent
D. 40 g water is fomed

Answer

To solve the problem, we need to analyze the reaction between hydrogen (H₂) and oxygen (O₂) to form water (H₂O). The balanced chemical equation for this reaction is:

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

### Step-by-Step Solution:

**Step 1: Determine the molar masses of the reactants.**
- Molar mass of H₂ = 2 g/mol (1 g/mol for H, and there are 2 H atoms)
- Molar mass of O₂ = 32 g/mol (16 g/mol for O, and there are 2 O atoms)

**Step 2: Calculate the number of moles of each reactant.**
- Moles of H₂ = \(\frac{8 \text{ g}}{2 \text{ g/mol}} = 4 \text{ moles}\)
- Moles of O₂ = \(\frac{32 \text{ g}}{32 \text{ g/mol}} = 1 \text{ mole}\)

**Step 3: Use the balanced equation to find the stoichiometric requirements.**
From the balanced equation, we see that:
- 2 moles of H₂ react with 1 mole of O₂.
- Therefore, 4 moles of H₂ would require \( \frac{4}{2} = 2 \text{ moles of O₂} \).

**Step 4: Compare the required moles of O₂ with the available moles.**
- Required moles of O₂ = 2 moles (for 4 moles of H₂)
- Available moles of O₂ = 1 mole

Since we need 2 moles of O₂ but only have 1 mole available, O₂ is the limiting reagent.

**Step 5: Determine the amount of water produced.**
According to the balanced equation:
- 2 moles of H₂ react with 1 mole of O₂ to produce 2 moles of H₂O.
- Since we have 1 mole of O₂, it will produce \(2 \text{ moles of H₂O}\).

**Step 6: Calculate the mass of water produced.**
- Molar mass of H₂O = 18 g/mol
- Moles of H₂O produced = 2 moles
- Mass of H₂O = \(2 \text{ moles} \times 18 \text{ g/mol} = 36 \text{ g}\)

### Conclusion:
- O₂ is the limiting reagent.
- H₂ is in excess.
- The mass of water produced is 36 g.

### Final Answer:
The correct statement is: "O₂ is the limiting reagent."