How many grams of KCLO₃ must be decomposed to prepare 3.36 litres of oxygen at STP ?

Yu dint explain nicely at all

No of moles = Given mass / Molar mass
The molar mass Of KClO3 =39+35.5+48=122.5
Also No of moles = given volume / 22.4

so, Given mass / Molar mass = Given volume / 22.4
        Given mass / 122.5 = 3.36 / 22.4
  
               Given mass= 3.36*122.5/22.4
                Mass required = 18.375 g


 

KClO3--- KCl + 3O2 247g 3x22.4L according to this equation,, 247g of KClO3 decompose to form 3x22.4L = 67.2L of O2 at STP => 67.2L of O2 is formed by KClO3 = 247g 3.36g of O2 will be formed by KClO3 = 247x3.36/67.2 =12.35g