Question

Small rocket engines, such as model rockets, contain KClO3 to provide oxygen for the combustion reaction fueling the rocket motor. The relevant reaction for KClO3 is: 2 KClO3(s) → 2 KCl(s) + 3 O2(g)

If 117.0 grams of solid KClO3 are burned in one of these rocket motors, how many liters of gaseous O2 are produced at 1.18 atm of pressure and 550.0 ºC?

Answer 1

moles of KClO3 = 117 / 122.55 = 0.9547

2 KClO3 (s) --------------> 2 KCl (s) + 3 O2 (g)

   2 mol                                                     3 mol

0.9547 mol                                                  ??

moles of O2 = 0.9547 x 3 / 2 = 1.432

moles of O2 = 1.432

pressure = 1.18 atm

temperature = 550. 0 oC = 823.15 K

P V = n R T

1.18 x V = 1.432 x 0.0821 x 823.15

V = 82.0 L

volume of O2 = 82.0 L