Question

Find out the number of molecules and volume of oxygen that evolves when 1225 g of KClO3 is heated

Answer 1

Answer:

A. 9.0345 x 10²⁴ molecules.

B. 336 liters

Explanation:

Calculate the Molar Mass of KClO₃:

Molar mass : Mass(K) + Mass(Cl) + Mass(O)

⇒ (1 x 39) + (1 x 35.5) + (3 x 16) = 39 + 35.5 + 48 = 122.5 grams per mole.

Given mass of KClO₃ = 1225 g

Number of moles of 1225 g of KClO₃ = (1225/122.5) = 10 moles.

We know that upon heating KClO₃, it decomposes to KCl and O₂

⇒ KClO₃ ------> KCl + $\frac{3}{2}$O₂

∴ 1 molecule of KClO₃ decomposes to 1 molecule of KCl and 1.5 moles of O₂.

∴ 122.5 g [1 mole] of KClO₃ will produce 1.5 moles of O₂ molecules.

∴ 1225 g [10 moles] of KClO₃, on complete decomposition, will produce,

⇒ 10 x 1.5 = 15 moles of O₂ molecules.

We know that 1 mole = 6.023 x 10²³ entities,

∴ 15 moles = 15 x 6.023 x 10²³ molecules = 9.0345 x 10²⁴ molecules of O₂.

∴ Upon complete decomposition of 1225 g of KClO₃, the number of O₂ molecules evolved is 9.0345 x 10²⁴ molecules.

We also know that volume of 1 mole of any gas at STP is 22.4 liters.

∴ Volume of O₂ gas evolved = Volume at STP x Number of moles of O₂

                                                = 22.4 x 15

                                                = 336 litres.

∴ Complete decomposition of 1225 g of KClO₃ produces 336 liters of O₂ gas.

Hope this helps!