• Data

Mass of Potassium chlorate = 15g
Molar mass of Potassium chlorate =122.5
Molar volume at S.T.P in 22.4dm3/mole

i. Mass of potassium chloride
ii. Volume of air (oxygen) formed
iii. Moles of air (oxygen) formed
• Balanced chemical equation

2KCLO3(s) 2KCL (2) + 3O2(g)
• Convert 15g into moles

Moles = mass / Molar mass
= 15g / 122.5g/mole
= 0.122 mole of KCLO3
• i / Find mass of Potassium Chloride produced

Make mole equivalent between KCLO3 and

2 moles of KCLO3 Ξ 2moles of KCL
0.1224489 mole of KLO3 Ξ ?? of KCL

(0.1224489 × 2) mole2 / 2mole
= 0.1224489 mole of KCL

Convert 0.1224489 moles of KCL into mass formula.
Mole = mass / Molar mass
Mass = mole × molar mass
= 0.1224489 moles × 74.55/mole
= 40.2755g

40.2755g of potassium chloride was formed
• ii/ Find the volume of oxygen as formed.

Use mole obtained from KCLO3 to relate with oxygen

2moles of KCLO3 Ξ 3moles of O2
0.1224489 moles of KCLO3 Ξ ?? OF O2

(0.1224489 moles × 3) / 2 moles
= 0.18367335 mole of O2

Convert 0.18367335 mole of O2 into volume at S.T.P (22.4dm3/mole)
Mole = volume / Molar volume at S.T.P
Molar volume at S.T.P = 22.4dm3/mole
Mole =0.18367335 molar

Volume = mole × molar volume at S.T.P
= 0.18367335 molar × 22.4dm3/mole
= 4.11428dm3

4.1dm3 of oxygen gas was produced S.T.P
• iii/ Calculate the moles of oxygen (air) formed

use mole equivalent between KCLO3 and O2

2 moles of KCLO3 Ξ 3 moles of O2
0.1224489 moles of KCLO3 Ξ ?? of O2
= (0.1224489 mole × 3 moles) / 2 moles

0.18367335 moles of oxygen (air) was formed.