Question

How many grams of KCl(s) are produced from the thermal decomposition of KClO₃(s) which produces 125.0 mL of O₂(g) at 25°C and 1.00 atm pressure according to the chemical equation shown below?
2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)?

Answer 1

we have:

P = 1.0 atm

V = 125.0 L

T = 25.0 oC

= (25.0+273) K

= 298 K

find number of moles using:

P * V = n*R*T

1 atm * 125 L = n * 0.08206 atm.L/mol.K * 298 K

n = 5.109 mol

This is number of moles of O2

From reaction,

moles of KCl produced = (2/3)*moles of O2

= (2/3)*5.109

= 3.406 mol

Molar mass of KCl = 1*MM(K) + 1*MM(Cl)

= 1*39.1 + 1*35.45

= 74.55 g/mol

we have below equation to be used:

mass of KCl,

m = number of mol * molar mass

= 3.406 mol * 74.55 g/mol

= 254 g

Answer: 254 g