PROBLEM 108P

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?

Solution 108P

Given that

Volume of oxygen = 235 mL.

Temperature (T) = 25 °C .

Pressure (P) = 697 torr .

Simply use the ideal gas equation: PV = nRT

where:

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