4.90g of KClO3,on heating ,shows a weight loss of .384g.what per cent of the orignal KClO3 has decomposed?

Dear student,

Decomposition=Initia-final/initial * 100

First step is to look for the balanced equation of the reaction.....2KCLO3---on heating------- 2KCL + 3 O2The weight of O2 decomposed is the loss in weight..The no. Of moles of O2 generated is =0.384/32=0.012 molesSince 2 moles of KCLO3 requires 3 moles of Oxygen ... Therefore....No. Of moles of KCLO3 produced is 0.012×(2/3)=0.008 molesTherefore the no. Of grams of KCLO3 used is :- 0.008×122.5= 0.98 gramsTherefore percentage of KCLO3 decomposed is ....(0.98/4.90)×100= 20percentThat`s the answer .....😊😊😊