Data

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

Asked:
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.