Answer to Question #70789 in General Chemistry for whitney

Question

1. You have a 0.5576 gram mixture of KClO3 and a non-reactive compound. You place the compound in our experimental setup and get the following results.

mass of O2 = 0.08420 g

a. If the mixture were pure KClO3, what is the theoretical yield of O2 that would be produced?

b. What is the mass percent of KClO3 in the mixture?

Answer 1

a) If the mixture were pure KClO3, m mixture = m KClO3 = 0.5576 gram
n KClO3 = m KClO3/ Mr KClO3=0.5576 g /122,55 g/mole= 0,005 mole
KClO 3 decomposition reaction is:
2 KClO 3(s) → 3 O 2(g) + 2 KCl (s)
Solve proportion :
0,005 mole KClO3 = X mole O2
2 mole KClO3 = 3 mole O2
X=n O2 from reaction is 0,0075 mole i.e. m O2= n O2* Mr O2 = 0,0075*32= 0,24 gramm
0.5576 gram of KClO3 will theoretically produce 0,24 gramm of O2
b) n O2= m O2/Mr O2 = 0.08420 g /32 g/mole = 0,0026 mole
2 KClO 3(s) → 3 O 2(g) + 2 KCl (s)
Solve proportion :
Y mole KClO3 = 0,0026 mole O2
2 mole KClO3 = 3 mole O2
Y=n KClO3 from reaction is 0,00173 mole i.e. m KClO3= n KClO3* Mr KClO3 = 0,00173*122,55=
0,21242 gramm
w (%)= m KClO3/m mixture X 100 % = 0,21242/0.5576 x 100% = 38,1 %
Answer: a)0,24 gramm O2 b) 38,1 % KClO3