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6. During an experiment, 17.5 mL of oxygen gas were collected over water at room temperature [tex]\(\left(25^{\circ} C \right)\)[/tex] and 100.2 kPa of atmospheric pressure.

a) What is the pressure of the "dry" oxygen gas?
[tex] P_T - P_{H_2O} = P_{\text{gas}} [/tex]
[tex] 100.2 \, \text{kPa} - 3.2 \, \text{kPa} = 97 \, \text{kPa} [/tex]

b) How many moles of oxygen gas were produced?
[tex] P V = n R T [/tex]
[tex] (97 \, \text{kPa})(0.0175 \, \text{L}) = n \left(0.08206 \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}}\right)(298 \, \text{K}) [/tex]

Note: Please ensure the unit conversions and constants are accurate when solving the equations.

Answer :

Sure! Let's solve this question step-by-step.

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Part a: Determining the Pressure of "Dry" Oxygen Gas

1. Identify the given values:
- Total atmospheric pressure, [tex]\( P_T = 100.2 \, \text{kPa} \)[/tex]
- Vapor pressure of water at [tex]\( 25^\circ \text{C} \)[/tex], [tex]\( P_{\text{H}_2\text{O}} = 3.2 \, \text{kPa} \)[/tex]

2. Calculate the pressure of the dry oxygen gas:
[tex]\[
P_{\text{gas}} = P_T - P_{\text{H}_2\text{O}} = 100.2 \, \text{kPa} - 3.2 \, \text{kPa} = 97.0 \, \text{kPa}
\][/tex]

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Part b: Calculating the Moles of Oxygen Gas Produced

1. Identify additional given values:
- Volume of oxygen gas, [tex]\( V = 17.5 \, \text{mL} = 0.0175 \, \text{L} \)[/tex] (converted from mL to L)
- Universal gas constant, [tex]\( R = 0.08206 \, \frac{\text{L atm}}{\text{mol K}} \)[/tex]
- Temperature in Celsius, [tex]\( 25^\circ \text{C} \)[/tex]

2. Convert temperature to Kelvin:
[tex]\[
T = 25 + 273.15 = 298.15 \, \text{K}
\][/tex]

3. Convert pressure from kPa to atm:
[tex]\[
P_{\text{gas\_atm}} = \frac{97.0 \,\text{kPa}}{101.325 \,\text{kPa/atm}} \approx 0.957 \, \text{atm}
\][/tex]

4. Use the ideal gas law to calculate the number of moles of oxygen gas:
[tex]\[
PV = nRT \implies n = \frac{PV}{RT}
\][/tex]
[tex]\[
n = \frac{(0.957 \, \text{atm})(0.0175 \, \text{L})}{(0.08206 \, \frac{\text{L atm}}{\text{mol K}})(298.15 \, \text{K})}
\][/tex]
[tex]\[
n \approx 0.000685 \, \text{moles}
\][/tex]

So, the final answers are:
- The pressure of the dry oxygen gas is [tex]\( 97.0 \, \text{kPa} \)[/tex].
- The amount of moles of oxygen gas produced is approximately [tex]\( 0.000685 \, \text{moles} \)[/tex].

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