how many moles of oxygen gas were evolved?

Question

A test tube containing a mixture of KCl, KClO₃, and MnO₂, having a total weight of 23.584g was heated to decompose the KClO_3. After heating, the mass was found to be 22.347g. 

a) How many moles of oxygen gas were evolved?

b) Calculate the number of moles and the mass of KClO₃ originally present in the test tube.

c) If the original mass of the sample was 4.39g, calculate the percent of KCl)₃ in the original sample. 

Answer 1

The balanced equation for the thermal decomposition of KClO3 is 2KClO3 ==> 2KCl + 3O2 

The MnO2 is a catalyst in this combustion reaction.

The only volatile product of the reaction is O2 gas. Thus,

23.584 g - 22.347 g = 1.237 g = mass lost = mass O2 

a) moles O2 = 1.237 g O2 x 1 mole/32 g = 0.03866 moles O2

b) moles KClO3 originally present = 0.03866 moles O2 x 2 moles KClO3/3 moles O2 = 0.02577 moles KClO3

mass KClO3 originally present = 0.02577 moles x 122.55 g/mole = 3.159 g KClO3

c) 3.159 g/4.39 g (x100%) = 71.9%