Question

How many grams of potassium chlorate decompose to form potassium chloride and 650. mL O2 at 128°C and 663 torr?

Answer 1

2 KClO3(s) ----> 2 KCl(s) + 3 O2(g)

use the ideal gas equation to calculate the no o fmoles O2

PV = nRT

P = 663 torr convert in to atm = 0.8723 atm

V = 650 mL = 0.65L

R = 0.0821 L atm/mol-k

T = 273 +128 = 401 k

put all these in the above equation

0.8723 x 0.65 = n x 0.0821 x 401K

n = 0.56699/ 32.9221

n = 0.01722 moles of O2

from the balanced equation 2 moles of KClO3 produces 3 moles of O2

how many moles of KClO3 from the 0.01722 moles of O2

= 0.01722 x 2/3

= 0.01148 moles of KClO3

Mass of ClO3 = moles x molar mass = 0.01488 x 122.54 = 1.40 grams