Which of the following statements `is//are` correct?
A mixture containing `64.0g H_(2)` and `64.0 g O_(2)` is ignited so that water is formed as follows:
`2H_(2) + O_(2) rarr 2H_(2) O`

Options

A. `H_(2) + O_(2) rarr 2 H_(2) O`
B. `O_(2)` is the limiting reagent
C. The reaction mixture contains `72.0 g of H_(2) O` and `56.0 g` of unreacted `H_(2)`.
D. The reaction mixture contains `56.0 g` of `H_(2) O` and `72.0 g` of unreacted `H_(2)`.

Explanation

To solve the question, we need to analyze the given reaction and the amounts of reactants involved. The reaction is:

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

### Step 1: Calculate the moles of \( H_2 \) and \( O_2 \)

1. **Calculate moles of \( H_2 \)**:
- Given mass of \( H_2 = 64.0 \, g \)
- Molar mass of \( H_2 = 2.0 \, g/mol \)
- Moles of \( H_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{64.0 \, g}{2.0 \, g/mol} = 32 \, moles \)

2. **Calculate moles of \( O_2 \)**:
- Given mass of \( O_2 = 64.0 \, g \)
- Molar mass of \( O_2 = 32.0 \, g/mol \)
- Moles of \( O_2 = \frac{64.0 \, g}{32.0 \, g/mol} = 2 \, moles \)

### Step 2: Determine the limiting reagent

From the balanced equation, we know:
- 1 mole of \( O_2 \) reacts with 2 moles of \( H_2 \).
- Therefore, 2 moles of \( O_2 \) would require \( 2 \times 2 = 4 \) moles of \( H_2 \).

Since we have:
- 32 moles of \( H_2 \) (which is in excess)
- 2 moles of \( O_2 \)

This indicates that \( O_2 \) is the limiting reagent because it will be consumed first in the reaction.

### Step 3: Calculate the amount of water produced

Using the stoichiometry of the reaction:
- 2 moles of \( O_2 \) produce 4 moles of \( H_2O \).
- Therefore, 2 moles of \( O_2 \) will produce \( 4 \, moles \, of \, H_2O \).

### Step 4: Calculate the mass of water produced

- Molar mass of \( H_2O = 18.0 \, g/mol \)
- Mass of \( H_2O = \text{moles} \times \text{molar mass} = 4 \, moles \times 18.0 \, g/mol = 72.0 \, g \)

### Step 5: Calculate the remaining amount of \( H_2 \)

Since 4 moles of \( H_2 \) are consumed:
- Initial moles of \( H_2 = 32 \, moles \)
- Moles of \( H_2 \) left = \( 32 - 4 = 28 \, moles \)

### Step 6: Calculate the mass of unreacted \( H_2 \)

- Mass of unreacted \( H_2 = 28 \, moles \times 2.0 \, g/mol = 56.0 \, g \)

### Conclusion

1. The limiting reagent is \( O_2 \).
2. The reaction mixture contains 72.0 g of \( H_2O \) and 56.0 g of unreacted \( H_2 \).

### Final Answer

The correct statements are:
- \( O_2 \) is the limiting reagent.
- The reaction mixture contains 72 g of \( H_2O \) and 56 g of unreacted \( H_2 \).