Question

Small rocket engines, such as model rockets, contain KClO3 to provide oxygen for the combustion reaction that fuels the rocket motor. The relevant reaction for KClO3 is: 2 KClO3(s) → 2 KCl(s) + 3 O2(g). If 141.6 grams of solid KClO3 are burned in one of these rocket motors, how many liters of gaseous O2 will be produced at 1.18 atm of pressure and 538oC?

Answer 1

first baance the equation

2KClO3(s) → 2 KCl(s) + 3 O2(g)

from this balanced equation it is clear that two moles of KClO3 is producing 3 moles of Oxygen

find out th eno of moles of KCLO3 from the weight

moles of KClO3 = weight of KClO3 / molar mass of KClO3

= 141.6 g / 122.5453 g /mol

= 1.155 mol

from balanced equation one mole of KClO3 is giving 3/2 moles of O2

accordingly

1.095 moles of KCLO3 will give 3/2 (1.155) = 1.733 moles of O2

now use the formula

PV = nRT

P is given as 1.18 atm, T = 538 ºC = 273 + 538 = 811K,

n = no of moles of O2 = 1.733 , R = 0.0821 Latm/mol-K

1.18 atm x V = 1.733 mol x 0.0821 Latm/mol-K x 811K

V = 97.78 L