Day 5: STP & Gas Stoich

Molar Volume at Standard Temperature and Pressure (STP)

  • **Definition of STP

    • STP stands for Standard Temperature and Pressure.**

    • It is defined as:

    • Standard Temperature: 0 degrees Celsius

    • Standard Pressure: 1 atmosphere

Calculating Molar Volume of a Gas at STP

  • Volume Calculation for One Mole of Gas at STP:

    • At STP, we can calculate the volume of one mole of a gas using the formula:

    • V=nRTPV = \frac{nRT}{P}

      • Where:

      • n = number of moles (which is 1 for one mole of gas)

      • R = Ideal Gas Constant (0.0821 L·atm/(K·mol))

      • T = temperature in Kelvin (0 degrees Celsius = 273 K)

      • P = pressure in atmospheres (1 atm)

    • Substituting the known values:

    • V=1×0.0821×2731=22.4 litersV = \frac{1 \times 0.0821 \times 273}{1} = 22.4 \text{ liters}

    • This tells us that one mole of any gas occupies a volume of 22.4 liters at STP.

Properties of Gas at STP

  • Behavior of Gases:

    • At STP, the identity of the gas does not matter; gas particles are so far apart that their size is negligible.

    • One mole of helium, xenon, methane, oxygen, or radon will all occupy 22.4 liters.

    • Thus, you can use the conversion factor:

    • (1 mole=22.4 liters)(1 \text{ mole} = 22.4 \text{ liters}) at STP.

Gas Volume Calculation Examples

  • Volume of Nitrogen:

    • For nitrogen:

    • extVolume=1 mole×22.4 liters/mole=22.4 litersext{Volume} = 1 \text{ mole} \times 22.4 \text{ liters/mole} = 22.4 \text{ liters}

    • For 2 moles of nitrogen:

    • 2×22.4 liters=44.8 liters2 \times 22.4 \text{ liters} = 44.8 \text{ liters}

    • Therefore, 1 mole of nitrogen = 22.4 liters and 2 moles = 44.8 liters.

  • Volume of Oxygen (from grams):

    • Given 24.8 grams of oxygen and using the molar mass (approximately 32 g/mole):

    • Calculate moles of oxygen:

      • n=24.8 grams32 grams/mole=0.775 molesn = \frac{24.8 \text{ grams}}{32 \text{ grams/mole}} = 0.775 \text{ moles}

    • Use conversion factor:

      • 0.775 moles×22.4 liters/mole≈17.4 liters0.775 \text{ moles} \times 22.4 \text{ liters/mole} \approx 17.4 \text{ liters}

  • Molecular Reference:

    • Given that 1 mole = 6.02×10236.02 \times 10^{23} molecules, thus:

    • 1.24 liters of hydrogen gas contains approximately:

      • 1.24 liters22.4 liters/mole×6.02×1023\frac{1.24 \text{ liters}}{22.4 \text{ liters/mole}} \times 6.02 \times 10^{23}

      • Results in approximately 1.20×10231.20 \times 10^{23} molecules of hydrogen.

Comparison of Gas Volumes at STP

  • Comparing 1 Gram Samples of Gases:

    • Different gases have different molar masses:

    • Hydrogen: Molar mass ≈ 2 g/mole

    • Oxygen: Molar mass ≈ 32 g/mole

    • Argon: Molar mass ≈ 40 g/mole

    • Greatest Volume Correlates with Greatest Molar Mass:

      • 1 gram of hydrogen has a larger number of moles (about 0.5 moles) compared to 1 gram of oxygen or argon, leading to a larger volume.

Ideal Behavior and Calculations

  • Calculating Water from Hydrogen:

    • For 1.24 liters of hydrogen gas, finding grams of water formed:

    • Use the reaction:

      • 2H2+O2→2H2O2H2 + O2 \rightarrow 2H_2O

    • From 1 mole of hydrogen, producing 1 mole of water (18.02 grams/mole) gives approximately 1 gram of water.

  • Finding Required Oxygen for Given Water Weight:

    • For 10.5 grams of water:

    • 10.5 grams18.02 grams/mole×1 mole O22 moles H2O×22.4 liters/mole\frac{10.5 \text{ grams}}{18.02 \text{ grams/mole}} \times \frac{1 \text{ mole O}2}{2 \text{ moles H}2O} \times 22.4 \text{ liters/mole}

    • Results in approximately 6.53 liters of oxygen.

Application of the Ideal Gas Law

  • Example with Non-STP Conditions:

    • Given 294 grams of potassium chlorate at a pressure and temperature, to find the liters of oxygen gas produced:

    • Step 1: Molar mass of potassium chlorate is about 122.55 g/mole.

    • Step 2: Convert grams to moles:

    • For every 2 moles of potassium chlorate, produce 3 moles of oxygen.

    • 294 grams122.55 grams/mole×3 moles O22 moles KClO3≈3.6 moles O2\frac{294 \text{ grams}}{122.55 \text{ grams/mole}} \times \frac{3 \text{ moles O}2}{2 \text{ moles KClO}3} \approx 3.6 \text{ moles O}_2 Calculating Volume of Oxygen Using Ideal Gas Law:

    • Given Pressure (converted from 70.75 mm Hg to atmospheres ≈ 0.98 atm) and Temp (305 K), calculate volume:

    • Using Ideal Gas Law:

    • V=nRTPV = \frac{nRT}{P}

    • Substitute values:

    • V=3.6×0.0821×3050.98V = \frac{3.6 \times 0.0821 \times 305}{0.98}

    • Results in approximately 90.8 liters of oxygen.

Last Example with Known Gas Volume

  • Using Ideal Gas Law for Given Volume of Oxygen:

    • For 4.58 liters of oxygen at a pressure of 745 mm Hg (converting to atmospheres yields 0.99 atm) and temperature of 308 K:

    • Use the Ideal Gas Law:

      • n=PVRTn = \frac{PV}{RT}

    • Substituting values yields approximately 0.177 moles of oxygen (4.58 liters).

    • Conversion to grams of silver oxide:

      • For every 1 mole of oxygen needed, 2 moles of silver oxide are required.

      • 0.177 moles O2×2 moles Ag2O1 mole O2×231.7 grams/mole Ag2O=82.0 grams of silver oxide0.177 \text{ moles O}2 \times \frac{2 \text{ moles Ag}2O}{1 \text{ mole O}2} \times 231.7 \text{ grams/mole Ag}2O = 82.0 \text{ grams of silver oxide}

  • Conclusion:

    • The study of the molar volume at STP is essential in understanding gas behavior, providing a conversion factor for various applications in chemistry and enhancing our understanding of ideal gas laws and stoichiometry.