10 g of hygrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaxtion will be:

Options

A. 2 moles
B. 3 moles
C. 4 moles
D. 1 moles

Explanation

To find the amount of water produced when 10 g of hydrogen reacts with 64 g of oxygen, we can follow these steps:

### Step 1: Write the Balanced Chemical Equation
The balanced chemical equation for the reaction between hydrogen and oxygen to form water is:
\[ 2H_2 + O_2 \rightarrow 2H_2O \]

### Step 2: Calculate Moles of Hydrogen and Oxygen
To determine the number of moles of each reactant, we use the formula:
\[ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

- **Molar mass of Hydrogen (H₂)** = 2 g/mol
- **Molar mass of Oxygen (O₂)** = 32 g/mol

**For Hydrogen:**
\[ \text{moles of } H_2 = \frac{10 \text{ g}}{2 \text{ g/mol}} = 5 \text{ moles} \]

**For Oxygen:**
\[ \text{moles of } O_2 = \frac{64 \text{ g}}{32 \text{ g/mol}} = 2 \text{ moles} \]

### Step 3: Determine the Limiting Reagent
From the balanced equation, we see that:
- 2 moles of \( H_2 \) react with 1 mole of \( O_2 \).

To find out how much \( O_2 \) is needed for 5 moles of \( H_2 \):
\[ \text{moles of } O_2 \text{ required} = \frac{5 \text{ moles of } H_2}{2} = 2.5 \text{ moles} \]

Since we only have 2 moles of \( O_2 \), oxygen is the limiting reagent.

### Step 4: Calculate the Amount of Water Produced
According to the stoichiometry of the balanced equation:
- 1 mole of \( O_2 \) produces 2 moles of \( H_2O \).

Thus, 2 moles of \( O_2 \) will produce:
\[ \text{moles of } H_2O = 2 \times 2 = 4 \text{ moles} \]

### Step 5: Conclusion
The amount of water produced in this reaction is 4 moles.

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