Question

Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO3 ( is ) . The equation for the reaction is 2 KClO3 ⟶ 2 KCl + 3 O 2 Calculate how many grams of O 2 ( g ) can be produced from heating 66.1 g KClO 3 ( s ) .

Answer 1

Answer: 25.89g of O2 (g)

Explanation:

We begin by finding the molar mass of compounds of interest in the given equation. 2 KClO3 ⟶ 2 KCl + 3 O 2

1  mole of KCLO3 = 39 .10 + 35.5 + (3x16) = 122.55g, 2Moles would be 2 x 122.5 = 245.1g

1 mole Oxygen gas = 32g, 3 moles would be, 3 x 32= 96g

According to the equation;

245.1g of KCLO3 produces 96g of O2 (g)

If 1g of KCLO3  produces  96÷245.1  of O2 (g)  from heating.

66.1g of KCLO3 would produce; 96÷245.1  x 66.1 = 25.89g of O2 (g) from heating.