Question

How many grams of potassium chlorate decompose to potassium chloride and 719 mL of O2 at 128°C and 685 torr?

2 KClO3(s) → 2 KCl(s) + 3 O2 (g)

*please explain

Answer 1

Given:

P = 685.0 torr

= (685.0/760) atm

= 0.901316 atm

V = 719.0 mL

= (719.0/1000) L

= 0.719 L

T = 128.0 oC

= (128.0+273) K

= 401 K

find number of moles using:

P * V = n*R*T

0.901316 atm * 0.719 L = n * 0.08206 atm.L/mol.K * 401 K

n = 1.968*10^-2 mol

This is number of moles of O2 formed

2 KClO3(s) → 2 KCl(s) + 3 O2 (g)

From balanced reaction above,

moles of KClO3 reacted = (2/3)*moles of O2 formed

= (2/3)*1.968*10^-2 mol

= 0.01312 mol

Molar mass of KClO3,

MM = 1*MM(K) + 1*MM(Cl) + 3*MM(O)

= 1*39.1 + 1*35.45 + 3*16.0

= 122.55 g/mol

mass of KClO3,

m = number of mol * molar mass

= 1.312*10^-2 mol * 122.55 g/mol

= 1.61 g

Answer: 1.61 g