2KClO3(s)= 2KCl(s) + 3O2(g)

Question

PLEASE do not put the answer only.

WRITE THE STEPS OF HOW YOU GET the answer with the LAWS!

2KClO3(s)= 2KCl(s) + 3O2(g)

The O₂ gas formed is collected over water in a flask where the total pressure is 624 mm Hg and the temperature is 16.0°C in an experiment, at which temperature the vapor pressure of water is 12 mm Hg.

What is the partial pressure of O₂ in the flask? ............ mm Hg.

If the wet O₂ gas occupies a volume of 6.33 L, how many moles of O₂ are formed? ........ mol

Answer 1

To find the partial pressure of O₂ gas that has been collected over water, we need to subtract the vapor pressure of water from the total pressure. This will then give us the pressure due to the O₂ gas.

624 mm Hg - 12 mm Hg = 612 mm Hg = Partial pressure of O₂

To find the moles of O₂ gas, we can use the Ideal Gas Law, PV = nRT

P = pressure = 612 mm Hg

V = volume 6.33 L

n = moles = ?

R = gas constant = 62.36 L-mm Hg/K-mol (you could use 0.0821 Latm/Kmol if you changed 612 mm to atm)

T = temperature in K = 16ºC + 273 = 289K

Solving for n (moles) we have...

n = PV/RT = (612)(6.33) / (62.36)(289)

n = 0.215 moles