Question

Potassium chlorate (KClO3) decomposes to form potassium chloride (KCl) and pure oxygen (O2) gas as shown below.

mc022-1.jpg

What amount of KClO3 is required to supply 28.0 L of oxygen gas at STP? (The molar mass of KClO3 is 122.55 g/mol.)
102 g
102.12 g
306.36 g
306.4 g

Answer 1

Are u sure these are the correct options?

Answer 2

yeah. i copied and pasted the question

Answer 3

Ok do u know the correct equation?

Answer 4

?

Answer 5

Because according to the equation I think 1 mol of KClO3 decompose to form 1 mol of KCl and 1.5 moles of oxygen so if 36dm3(1.5 into 24) of oxygen gives 122.55g of KClO3 so as 1 litre=1 dm3 so 28 litre=28 dm3 when cross multiplied gives 95.317 g of KClO3