Answer:
There is 2.52 L of O2 collected
Explanation:
Step 1: Data given:
Temperature = 22.0 °C
Pressure = 728 mmHg = 728 /760 = 0.958 atm
Mass of KClO3 = 8.15 grams
Molar mass of KClO3 = 122.55 g/mol
Step 2: The balanced equation
2KClO3(s) → 2KCl(s) + 3O2(g)
Step 3: Calculate moles of KClO3
Moles KClO3 = mass KClO3 / molar mass KClO3
Moles KClO3= 8.15 grams / 122.55 g/mol
Moles KClO3 = 0.0665 moles
Step 4: Calculate moles of O2
For 2 moles of KClO3 we'll have 2 moles of KCl and 3 moles of O2 produced
For 0.0665 moles of KClO3 we have 3/2 * 0.0665 = 0.09975 moles
Step 5: Calculate vlume of O2
p*V = n*R*T
V = (n*R*T)/p
⇒ with n = the number of moles O2 = 0.09975 moles
⇒ with R = the gas constant = 0.08206 L*atm/K*mol
⇒ with T = 22.0 °C = 273 +22 = 295 Kelvin
⇒ with p = 0.958 atm
V = (0.09975 * 0.08206 * 295) / 0.958
V = 2.52 L
There is 2.52 L of O2 collected
