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 determining the volume of gaseous products after the explosion of 10 g of hydrogen and 64 g of oxygen, we can follow these steps:

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

### Step 2: Calculate the molar masses
- Molar mass of \( H_2 \) (hydrogen) = 2 g/mol
- Molar mass of \( O_2 \) (oxygen) = 32 g/mol

### Step 3: Calculate the number of moles of reactants
- Moles of \( H_2 \):
\[ \text{Moles of } H_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{10 \, \text{g}}{2 \, \text{g/mol}} = 5 \, \text{moles} \]

- Moles of \( O_2 \):
\[ \text{Moles of } O_2 = \frac{64 \, \text{g}}{32 \, \text{g/mol}} = 2 \, \text{moles} \]

### Step 4: Determine the limiting reactant
From the balanced equation, we see that 2 moles of \( H_2 \) react with 1 mole of \( O_2 \). Therefore:
- For 2 moles of \( O_2 \), we need:
\[ 2 \times 2 = 4 \, \text{moles of } H_2 \]
Since we have 5 moles of \( H_2 \) available, \( O_2 \) is the limiting reactant.

### Step 5: Calculate the amount of \( H_2 \) that reacts
Since 2 moles of \( O_2 \) will react with 4 moles of \( H_2 \), and we have 2 moles of \( O_2 \):
- Moles of \( H_2 \) that react:
\[ 4 \, \text{moles of } H_2 \text{ (needed)} \text{ with } 2 \, \text{moles of } O_2 \]

### Step 6: Calculate the remaining \( H_2 \)
Initially, we had 5 moles of \( H_2 \):
- Moles of \( H_2 \) remaining:
\[ 5 - 4 = 1 \, \text{mole} \]

### Step 7: Calculate the volume of the gaseous products
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of the remaining \( H_2 \):
\[ \text{Volume of remaining } H_2 = 1 \, \text{mole} \times 22.4 \, \text{L/mole} = 22.4 \, \text{L} \]

### Final Answer
The volume of the gaseous product after the reaction is 22.4 liters of unreacted hydrogen.

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