CHEMISTRY | MOLE CONCEPT | Question 21
Calculate the volume of oxygen gas product at STP when 12.25g of potassium chlorate are decomposed by heating according to the equation below.
2KCLO3(S) heat→ 2KCL(S) + 302(S)
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DATA
Mass of KClOg = 12.25g
Molar volume = 22.4dm3
Equation
2KCLO3 (s) → 2KCl(s) + 3O2(g)
Volume of oxygen produced
Mole for KCLO3 = mass / Molar mass
= 12.25g / 122.5g/mole
To make mole equivalent between KCLO3 and O2
2 moles of KCLO3 Ξ 3 mole O2
0.1 mole of KCLO3 Ξ ?
= (0.1 mole × 3 moles) / 2 moles
= 0.15 mole
Volume = 0.15 mole × 22.4
= 3.36dm3
3.36dm3 of oxygen gas was produced
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