How many grams of oxygen are collected in a reaction where 230. mL of oxygen gas is collected over water at a temperature of 25°C and a total pressure of 697 torr?
To determine the number of grams of oxygen collected in the reaction, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, we need to convert the given conditions to the appropriate units.
The volume of oxygen gas, given as 230. mL, needs to be converted to liters:
230 mL / 1000 mL/L = 0.230 L
The temperature, given as 25°C, needs to be converted to Kelvin:
25°C + 273.15 = 298.15 K
The total pressure, given as 697 torr, needs to be converted to atm:
697 torr / 760 torr/atm = 0.917 atm
Now we can substitute these values into the ideal gas law equation:
0.917 atm * 0.230 L = n * 0.0821 L·atm/mol·K * 298.15 K
Simplifying the equation:
0.21071 = n * 24.456615
Now we solve for n (number of moles):
n = 0.21071 / 24.456615
n = 0.00862 mol
Finally, we can calculate the number of grams using the molar mass of oxygen (O2), which is approximately 32 g/mol:
Mass = n * molar mass
Mass = 0.00862 mol * 32 g/mol
Mass = 0.275 g
Therefore, approximately 0.275 grams of oxygen are collected in the reaction.
To calculate the number of grams of oxygen collected in this reaction, we need to use the ideal gas law equation, which is:
PV = nRT
Where:
P = pressure of the gas (in atm)
V = volume of the gas (in liters)
n = number of moles
R = ideal gas constant (0.0821 L∙atm/moles∙K)
T = temperature of the gas (in Kelvin)
First, we need to convert the volume of oxygen gas collected to liters. There are 1000 mL in a liter, so we divide 230 mL by 1000:
230 mL ÷ 1000 = 0.230 L
Now, let's convert the temperature from Celsius to Kelvin. To do this, we add 273 to the given temperature of 25°C:
25°C + 273 = 298 K
Next, we convert the total pressure from torr to atmospheres. There are 760 torr in 1 atm, so we divide the given pressure of 697 torr by 760:
697 torr ÷ 760 torr/atm = 0.917 atm
Now we have all the necessary information to solve the ideal gas law equation.
PV = nRT
(0.917 atm)(0.230 L) = n(0.0821 L∙atm/moles∙K)(298 K)
Now we can solve for the number of moles (n):
(0.917 atm)(0.230 L) = n(0.0821 L∙atm/moles∙K)(298 K)
0.211 cm^3 atm = n(24.4538 L∙atm/moles)
0.211 cm^3 atm ÷ 24.4538 L∙atm/moles = n
n ≈ 0.0086 moles
Finally, to calculate the number of grams of oxygen, we need to multiply the number of moles by the molar mass of oxygen. The molar mass of oxygen is 32.00 g/mol:
0.0086 moles × 32.00 g/mol = 0.2752 grams
Therefore, approximately 0.2752 grams of oxygen are collected in this reaction.