Answer:
The mass of oxygen collected is 0.0928 grams
Explanation:
<u>Step 1: </u>Data given
Yield of oxygen gas = 75.3 mL
Temperature = 25°C
Pressure = 742 torr = 0.976 atm
vapor pressure of water at 25 °C = 23.7695 torr = 0.0313 atm
Molecular weight of O2 = 32 g/mol
universal gas constant = 0.08206 L* atm / K* mol
<u>Step 2</u>: Calculate the vapor pressure (= pO2)
Ptotal + pO2 + pH2O
pO2 = ptotal -pH2O
pO2 = 0.976 atm - 0.0313 atm = 0.9447 atm
<u>Step 3</u>: Calculate moles O2:
p*V = n*R*T
⇒ p = the pressure of O2 = 0.9447 atm
⇒ V = the volume = 0.0753 L
⇒ n = the number of moles = TO BE DETERMINED
⇒ R= the gas constant = 0.08206 L*atm/K*mol ( 24 torr)
⇒ T = the temperature = 25°C = 298.15 Kelvin
n = (p*V)/(R*T)
n = (0.9447 atm * 0.0753 L) /(0.08206 l*atm/K*mol * 298.15 K)
number of moles = 0.0029 moles O2
<u>Step 4:</u> Calculate mass of O2
Mass O2 = moles O2 * Molar mass O2
Mass O2 = 0.0029 * 32 g/mol
Mass O2 =0.0928 grams
The mass of oxygen collected is 0.0928 grams
