Question

Compute the mass in grams of KClO3 necessary to produce 67.2litres of oxygen at STP according to the reaction
2KClO3-------- 2KCl + 3O2

Answer 1

2KClO3-------- 2KCl + 3O2

I mole of O2 = 32g = 22.4L

1L = 32/22.4 g

67.2L = 32 x 67.2 / 22.4 = 96g

3 moles (96g) of O2 require 2 moles of KClO3

102.5 x 2 = 205g of KClO3

Answer 2

Answer:

205g of the KClO3.

Explanation:

We know that the equation of the KClO3 dissociating into KCl and O2 which can be 2KClO3 -> 2KCl + 3O2.

So, we get that 1 mole of O2 is having 32g = present in 22.4L.

So, correspondingly for 1L we will have = 32/22.4 g  of O2.

So, from the question we have 67.2L which will be = 32 x 67.2 / 22.4 = 96g.

Since from the stoichiometric coefficient we get that 3 moles (96g) of O2 which will be requires the 2 moles of KClO3 will have mass of:-

102.5 x 2 = 205g mass of KClO3.