8. For the reaction 2H₂O(1) + 2e → H₂(g) + 2OH(aq), calculate the volume of "dry" hydrogen gas created at a pressure of 745 mm Hg and 25.0 °C when 0.6696 g H₂O are used. The vapor pressure of water at this temperature is 23.8 mmHg. A) 0.479 L B) 0.464 L C) 0.450 L D) 4.18 L E) 4.05 L

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Chapter1: Chemical Foundations
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**Problem Statement:**

For the reaction 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq), calculate the volume of "dry" hydrogen gas created at a pressure of 745 mm Hg and 25.0 °C when 0.6696 g H₂O are used. The vapor pressure of water at this temperature is 23.8 mmHg.

**Options:**

A) 0.479 L  
B) 0.464 L  
C) 0.450 L  
D) 4.18 L  
E) 4.05 L

**Detailed Explanation:**

To solve this problem, you need to determine the amount of hydrogen gas produced and then calculate the volume using the ideal gas law.

1. **Determine Moles of Water (H₂O) Used:**

Molar mass of H₂O = 2(1.008) + 16.00 = 18.016 g/mol  
Moles of H₂O = 0.6696 g / 18.016 g/mol = 0.0372 mol

2. **Calculate Moles of Hydrogen Gas (H₂) Produced:**

From the balanced chemical equation, 2 moles of H₂O produce 1 mole of H₂.  
Therefore, moles of H₂ = 0.0372 mol H₂O * (1 mol H₂ / 2 mol H₂O) = 0.0186 mol H₂

3. **Calculate Pressure of "Dry" Hydrogen Gas:**

Total pressure = 745 mm Hg  
Vapor pressure of water = 23.8 mm Hg  
Partial pressure of H₂ = 745 mm Hg - 23.8 mm Hg = 721.2 mm Hg

4. **Convert Pressure to Atmospheres:**

Pressure in atm = 721.2 mm Hg * (1 atm / 760 mm Hg) = 0.949 atm

5. **Use Ideal Gas Law (PV = nRT):**

Where:  
P = pressure in atm (0.949 atm)  
V = volume in liters (unknown)    
n = moles of gas (0.0186 mol)  
R = ideal gas constant (0.0821 L·atm/(K·mol))  
T = temperature in Kelvin (25.0 °C + 273.15 =
Transcribed Image Text:**Problem Statement:** For the reaction 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq), calculate the volume of "dry" hydrogen gas created at a pressure of 745 mm Hg and 25.0 °C when 0.6696 g H₂O are used. The vapor pressure of water at this temperature is 23.8 mmHg. **Options:** A) 0.479 L B) 0.464 L C) 0.450 L D) 4.18 L E) 4.05 L **Detailed Explanation:** To solve this problem, you need to determine the amount of hydrogen gas produced and then calculate the volume using the ideal gas law. 1. **Determine Moles of Water (H₂O) Used:** Molar mass of H₂O = 2(1.008) + 16.00 = 18.016 g/mol Moles of H₂O = 0.6696 g / 18.016 g/mol = 0.0372 mol 2. **Calculate Moles of Hydrogen Gas (H₂) Produced:** From the balanced chemical equation, 2 moles of H₂O produce 1 mole of H₂. Therefore, moles of H₂ = 0.0372 mol H₂O * (1 mol H₂ / 2 mol H₂O) = 0.0186 mol H₂ 3. **Calculate Pressure of "Dry" Hydrogen Gas:** Total pressure = 745 mm Hg Vapor pressure of water = 23.8 mm Hg Partial pressure of H₂ = 745 mm Hg - 23.8 mm Hg = 721.2 mm Hg 4. **Convert Pressure to Atmospheres:** Pressure in atm = 721.2 mm Hg * (1 atm / 760 mm Hg) = 0.949 atm 5. **Use Ideal Gas Law (PV = nRT):** Where: P = pressure in atm (0.949 atm) V = volume in liters (unknown) n = moles of gas (0.0186 mol) R = ideal gas constant (0.0821 L·atm/(K·mol)) T = temperature in Kelvin (25.0 °C + 273.15 =
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