10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Volume of gaseous product after reaction

Options

A. `1 xx 22.4L`
B. `2 xx 22.4L`
C. `3 xx 22.4L`
D. `4 xx 22.4L`

Answer

To solve the problem of finding the volume of gaseous products after the reaction of hydrogen and oxygen, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between hydrogen and oxygen can be represented as:
\[ 2H_2 + O_2 \rightarrow 2H_2O \]

### Step 2: Calculate the number of moles of hydrogen and oxygen
Using the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]

- For hydrogen (H₂):
- Given mass = 10 g
- Molar mass of H₂ = 2 g/mol
\[ \text{Number of moles of } H_2 = \frac{10 \, \text{g}}{2 \, \text{g/mol}} = 5 \, \text{moles} \]

- For oxygen (O₂):
- Given mass = 64 g
- Molar mass of O₂ = 32 g/mol
\[ \text{Number of moles of } O_2 = \frac{64 \, \text{g}}{32 \, \text{g/mol}} = 2 \, \text{moles} \]

### Step 3: Determine the limiting reagent
To find the limiting reagent, we compare the mole ratio from the balanced equation:

From the balanced equation:
- 2 moles of H₂ react with 1 mole of O₂.

Calculating the required moles:
- For 5 moles of H₂, the required moles of O₂:
\[ \text{Required moles of } O_2 = \frac{5 \, \text{moles of } H_2}{2} = 2.5 \, \text{moles} \]

Since we only have 2 moles of O₂ available, O₂ is the limiting reagent.

### Step 4: Calculate the amount of water produced
According to the balanced equation:
- 1 mole of O₂ produces 2 moles of H₂O.

Thus, from 2 moles of O₂:
\[ \text{Moles of } H_2O = 2 \, \text{moles of } O_2 \times 2 = 4 \, \text{moles of } H_2O \]

### Step 5: Calculate the volume of the gaseous product (water vapor)
Using the molar volume of a gas at standard temperature and pressure (STP), which is 22.4 L/mol:
\[ \text{Volume of } H_2O = \text{Number of moles} \times \text{Molar volume} \]
\[ \text{Volume of } H_2O = 4 \, \text{moles} \times 22.4 \, \text{L/mol} = 89.6 \, \text{L} \]

### Final Answer
The volume of the gaseous product (water vapor) after the reaction is **89.6 L**.

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