Question

How many litres (L) of oxygen at STP can be obtained from 110 g of potassium chlorate? [R=0.0821 L.atm/K.mol]

Answer 1

Balance the equation first:

2 KClO3 (s) ---> 2 KCl (s) + 3 O2 (g)

Moles of KClO3 =  110 / 122.5 = 0.89

Following the balanced chemical equation:
We can say moles of O2 produce =  $\frac{3}{2}$ x moles of KClO3

So, O2 = (3 / 2) x  0.89

= 1.34 moles

So, Volume at STP = nRT / P

T = 273.15 K
P = 1 atm

So, V = (1.34 x 0.0821 x 273.15) / 1  =  30.2 L