Question

How many grams of potassium chlorate decompose to form potassium chloride and 415 mL O₂ at 128°C and 686 torr?

g

Answer 1

balanced equation is

2 KClO3 -----> 2 KCl + 3O2

PV = nRT

P = pressure = 686 torr = 0.903 atm

V = volume = 415 mL = 0.415 L

T = temperature = 273 + 128 = 401 K

0.903*0.415 = n*0.0821*401

0.375 = n*32.9

n = 0.0114 mole of O2

from the balanced equation we can say that

3 mole of O2 is produced by 2 mole of potassium chlorate

so 0.0114 mole of O2 will produce 0.0076 mole of potassium chlorate

1 mole of potassium chlorate = 122.55 g

0.0076 mole of potassium chlorate = 0.931 g

Therefore, the mass of potassium chlorate = 0.931 g