Question

Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO3(s) . The equation for the reaction is 2KClO3⟶2KCl+3O2 Calculate how many grams of O2(g) can be produced from heating 70.4 g KClO3(s)

STRATEGY: Convert the mass of KClO3 to moles. Convert the number of moles of KClO3 to the number of moles of O2 . Convert the number of moles of O2 to grams. Step 1: 70.4 g KClO3 is equal to 0.574 mol KClO3 . Step 2: If 0.574 mol KClO3 reacted, how much O2 was produced?

Answer 1

Molar mass of KClO3,

MM = 1*MM(K) + 1*MM(Cl) + 3*MM(O)

= 1*39.1 + 1*35.45 + 3*16.0

= 122.55 g/mol

mass of KClO3 = 70.4 g

mol of KClO3 = (mass)/(molar mass)

= 70.4/1.226*10^2

= 0.5745 mol

Balanced chemical equation is:

2KClO3⟶2KCl+3O2

According to balanced equation

mol of O2 formed = (3/2)* moles of KClO3

= (3/2)*0.5745

= 0.8617 mol

Molar mass of O2 = 32 g/mol

mass of O2 = number of mol * molar mass

= 0.8617*32

= 27.57 g

Answer: 27.6 g