How many grams of oxygen are collected in a reaction where 245 mL of oxygen gas is collected over water at a temperature of 25 ^\circ {\rm C} and a total pressure of 688 torr
Look up vapor pressure of H2O at 25C and
subtract from 688 torr.
688 torr - vpH2O = pressure dry gas.
Then use PV = nRT and solve for n = number of mols oxygen. Don't forget P must be in atmospheres (torr dry gas/760) and T must be in kelvin. V must be in L.
Then n = grams O2/molar mass O2. Solve for grams.
To determine the number of grams of oxygen collected, we need to use the ideal gas law. The ideal gas law equation is:
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 of the gas (in Kelvin)
First, we need to convert the given values to the appropriate units:
- The temperature is in degrees Celsius, so we need to convert it to Kelvin. We add 273.15 to the given temperature:
T = 25°C + 273.15 = 298.15 K
- The pressure is given in torr, and we need to convert it to atm. Since 1 atm = 760 torr, we divide the given pressure by 760:
P = 688 torr / 760 torr/atm = 0.905 atm
Now we can plug the values into the ideal gas law equation to calculate the number of moles (n) of the gas released:
PV = nRT
(0.905)(V) = n(0.0821)(298.15)
Since the volume is given as 245 mL, which is equivalent to 0.245 L, we can substitute the values:
(0.905)(0.245) = n(0.0821)(298.15)
Solving for n:
0.221725 = 24.460415n
n ≈ 0.00906 moles
To find the mass of oxygen collected, we need to use its molar mass. The molar mass of oxygen (O2) is approximately 32 g/mol.
Using the formula mass = moles × molar mass:
mass = 0.00906 moles × 32 g/mol
mass ≈ 0.2907 grams
Therefore, approximately 0.2907 grams of oxygen are collected in this reaction.