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 66.8 g KClO 3 ( s ) .

The answer to your question is  52.35 g of O₂

Explanation:

mass of oxygen = ?

mass of KClO₃ = 66.8 g

Chemical balanced reaction

              KClO₃   ⇒   2 KCl  +   3O₂

Process

1.- Calculate the molar mass of KClO₃ and Oxygen

KClO₃ = 39 + 35.5 + (16 x 3) = 122.5 g

O₂ = 16 x 6 = 96 g

2.- Write proportions to solve this problem

                     122.5 g of KClO₃  96 g of O₂

                        66.8 g                x

                        x = (66.8 x 96) / 122.5

                        x = 6412.8/122.5

                        x = 52.35 g of O₂

hi,

the answer is d.indefinite shape but definite volume, let go oogle u out

hope this !

your answer would be a. 12.5 kg