To determine the pressure of the hydrogen gas produced in this reaction, we need to follow a series of steps involving stoichiometry and the Ideal Gas Law.
### Step 1: Understanding the Reaction
The balanced chemical equation for the reaction is:
[tex]\[ 2 \, \text{K(s)} + 2 \, \text{H}_2\text{O(l)} \rightarrow 2 \, \text{KOH(aq)} + \text{H}_2\text{(g)} \][/tex]
According to the balanced equation, 2 moles of potassium (K) produce 1 mole of hydrogen gas (H[tex]\(_2\)[/tex]).
### Step 2: Calculate Moles of Hydrogen Gas Produced
We have a 0.684 mol sample of potassium (K). Using the stoichiometry of the reaction:
[tex]\[ 2 \, \text{moles of K} \rightarrow 1 \, \text{mole of H}_2 \][/tex]
[tex]\[ \text{Therefore, 0.684 moles of K} \rightarrow \frac{0.684}{2} \, \text{moles of H}_2 \][/tex]
[tex]\[ \text{Moles of H}_2 = 0.342 \, \text{mol} \][/tex]
### Step 3: Use the Ideal Gas Law to Find the Pressure
The Ideal Gas Law equation is:
[tex]\[ PV = nRT \][/tex]
where:
- [tex]\( P \)[/tex] is the pressure,
- [tex]\( V \)[/tex] is the volume (3.00 L given),
- [tex]\( n \)[/tex] is the number of moles of hydrogen gas (0.342 moles),
- [tex]\( R \)[/tex] is the Ideal Gas Constant (0.0821 L·atm/(K·mol)),
- [tex]\( T \)[/tex] is the temperature (285 K).
Rearranging for pressure [tex]\( P \)[/tex]:
[tex]\[ P = \frac{nRT}{V} \][/tex]
### Step 4: Substitute the Values into the Ideal Gas Law
Substituting the known values into the formula:
[tex]\[ P = \frac{0.342 \times 0.0821 \times 285}{3.00} \][/tex]
[tex]\[ P = \frac{7.977}{3.00} \][/tex]
[tex]\[ P \approx 2.67 \, \text{atm} \][/tex]
### Conclusion
The pressure of the hydrogen gas inside the container is approximately 2.67 atm. So, the correct answer is:
[tex]\[ \boxed{2.67 \, \text{atm}} \][/tex]
