How many grams of oxygen are collected in a reaction where 235 mL of oxygen gas is collected over water at a temperature of 25 °C and a total pressure of 697 torr?
Volume of oxygen = 235 mL.
Temperature (T) = 25 °C .
Pressure (P) = 697 torr .
Simply use the ideal gas equation: PV = nRT
P = pressure = 717 torr
V = volume = 235 mL or 0.235 L
n = unknown
R = gas constant = 0.08206 L*atm/mol*K
T = temperature = 25 C + 273.15 = 298.15 K
Since the gas constant R is expressed in L*atm/mol K, the pressure should also be expressed in atm. Convert 697 torr to atm noting that 1 atm = 760 torr.
697*1 atm /760 atm = 0.917 atm .
Now substitute these values to the IDG equation to find the amount of substance (in moles)
0.9171*0.235 =n*(0.08206 L*atm/mol*K)*(298.15K)
n= 113 .53 moles .
To determine the mass of oxygen from the obtained result,
1 mole O2 = 2(atomic mass of oxygen) = 2(16.0g) = 32 g .
n = number of moles O2 = 8.74e-3 mol
8.74e-3 mol O2 * (32.00g O2 / 1mol O2) = .280g O2