Hydrogen produced from a hydrolysis reaction was collected over water. The data is compiled in the table. Total volume of H₂(g) collected 96.17 mL Temperature 27.0 °C Barometric pressure 738 mmHg Vapor pressure of water at 27.0 °C 27.0 mmHg Calculate the moles of hydrogen gas produced by the reaction. moles: mol H₂

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### Calculation of Moles of Hydrogen Gas Produced from Hydrolysis Reaction

Hydrogen produced from a hydrolysis reaction was collected over water. The data is compiled in the table below:

|                                 |                                  |
|---------------------------------|----------------------------------|
| **Total volume of H₂(g) collected**        | 96.17 mL                         |
| **Temperature**                        | 27.0 °C                          |
| **Barometric pressure**                | 738 mmHg                         |
| **Vapor pressure of water at 27.0 °C** | 27.0 mmHg                        |

**Task:**
Calculate the moles of hydrogen gas produced by the reaction.

**Formula:**
To calculate the moles of hydrogen gas (n), you can use the Ideal Gas Law equation:
\[ PV = nRT \]

Where:
- \( P \) is the pressure of the gas (in atm)
- \( V \) is the volume of the gas (in liters)
- \( n \) is the number of moles of the gas
- \( R \) is the ideal gas constant (0.0821 L·atm/(mol·K))
- \( T \) is the temperature (in Kelvin)

**Data given:**
- Total volume of H₂(g) collected = 96.17 mL = 0.09617 L (converted to liters)
- Temperature = 27.0 °C = 300.15 K (converted to Kelvin)
- Barometric pressure = 738 mmHg
- Vapor pressure of water at 27.0 °C = 27.0 mmHg

**Steps:**

1. **Calculate the partial pressure of H₂ gas:**
   \[
   P_{H₂} = \text{Barometric pressure} - \text{Vapor pressure of water}
   \]
   \[
   P_{H₂} = 738 \, \text{mmHg} - 27.0 \, \text{mmHg} = 711 \, \text{mmHg}
   \]
   Convert mmHg to atm:
   \[
   P_{H₂} = 711 \, \text{mmHg} \times \frac{1 \, \text{atm}}{760 \, \text{mmHg}} \approx 0.9355 \, \text{atm}
   \]

2. **Apply the Ideal Gas
Transcribed Image Text:### Calculation of Moles of Hydrogen Gas Produced from Hydrolysis Reaction Hydrogen produced from a hydrolysis reaction was collected over water. The data is compiled in the table below: | | | |---------------------------------|----------------------------------| | **Total volume of H₂(g) collected** | 96.17 mL | | **Temperature** | 27.0 °C | | **Barometric pressure** | 738 mmHg | | **Vapor pressure of water at 27.0 °C** | 27.0 mmHg | **Task:** Calculate the moles of hydrogen gas produced by the reaction. **Formula:** To calculate the moles of hydrogen gas (n), you can use the Ideal Gas Law equation: \[ PV = nRT \] Where: - \( P \) is the pressure of the gas (in atm) - \( V \) is the volume of the gas (in liters) - \( n \) is the number of moles of the gas - \( R \) is the ideal gas constant (0.0821 L·atm/(mol·K)) - \( T \) is the temperature (in Kelvin) **Data given:** - Total volume of H₂(g) collected = 96.17 mL = 0.09617 L (converted to liters) - Temperature = 27.0 °C = 300.15 K (converted to Kelvin) - Barometric pressure = 738 mmHg - Vapor pressure of water at 27.0 °C = 27.0 mmHg **Steps:** 1. **Calculate the partial pressure of H₂ gas:** \[ P_{H₂} = \text{Barometric pressure} - \text{Vapor pressure of water} \] \[ P_{H₂} = 738 \, \text{mmHg} - 27.0 \, \text{mmHg} = 711 \, \text{mmHg} \] Convert mmHg to atm: \[ P_{H₂} = 711 \, \text{mmHg} \times \frac{1 \, \text{atm}}{760 \, \text{mmHg}} \approx 0.9355 \, \text{atm} \] 2. **Apply the Ideal Gas
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