Question

PLEASE HELP!
The following equation is one way to prepare oxygen in a lab.

2KClO3 → 2KCl + 3O2

Molar mass Info: MM O2 = 32 g/mol MM KCl = 74.55 g/mol MM KClO3 = 122.55 g/mol


If 25.6 g of KCl are produced, then how many moles of KClO3 were consumed?

A. 0.209 mol
B. 0.687 mol
C. 0.343 mol
D. 0.172 mol

Answer 1

Molar mass:

KCl = 74.55 g/mol

KClO3 = 122. 55 g/mol

  Calculation of the mass of KClO3 :

2 KClO3 =  2 KCl + 3 O2

2* 122.55 g KClO3 ------------------ 2 * 74.55 g KCl
mass KClO3 ?? --------------------- 25.6 g KCl

mass KClO3 = 25.6 * 2 * 122.55 / 2 * 74.55

mass KClO3 = 6274.56 / 149.1

mass = 42.082 g of  KClO3

Therefore:

1 mole KClO3 ---------------------- 122.55 g
?? moles KClO3 ------------------- 42.082 g

moles KClO3 = 42.082 * 1 / 122.55

moles KClO3 = 42.082 / 122.55

=> 0.343 moles of KClO3


Answer C

hope this helps!