What is the mass of KClO _ 3( ~s ) required to liberate 22. 4 dm ^3 oxygen at STP during thermal decomposition? (Molar Mass of KClO _ 3( ~s ) =122.5 ~g / mol )

Step-by-step solution

aligned & 2 KClO ₃ & [2 moles ] aligned 2 KCl + [3 moles ] 3 O ₂ [3 moles] 2 moles of KClO ₃=2 122.5=245 ~g 3 moles of O ₂ at STP occupy = (3 22.4 dm ^3 ) . Thus, 245 ~g of potassium chlorate will liberate 67.2 dm ^3 of oxygen gas. Let ' x ' gram of KClO ₃ liberate 22.4 dm ^3 of oxygen gas at S.T.P. x= 245 22.4 3 22.4 =81.67 ~g