Question

How many grams of KCl(s) are produced from the thermal decomposition of KClO3(s) which produces 50.0 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)?

0.102 g

0.167 g

0.152 g

0.304 g

Answer 1

Given:

P = 1.0 atm

V = 50.0 mL

= (50.0/1000) L

= 0.05 L

T = 25.0 oC

= (25.0+273) K

= 298 K

find number of moles using:

P * V = n*R*T

1 atm * 0.05 L = n * 0.08206 atm.L/mol.K * 298 K

n = 2.045*10^-3 mol

Now use:

mol of KCl produced = (2/3)*number of mol of O2

= (2/3)*2.045*10^-3 mol

=1.363*10^-3 mol

Molar mass of KCl,

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

= 1*39.1 + 1*35.45

= 74.55 g/mol

use:

mass of KCl,

m = number of mol * molar mass

= 1.363*10^-3 mol * 74.55 g/mol

= 0.102 g

Answer: 0.102 g