In the reaction 2 KClO _ 3( ~s ) 2 KCl _ ( s ) +3 O _ 2( ~g ) H ^ =-78 ~kJ . If 33 6 ~L of oxygen gas is liberated at S.T.P. What is the mass of KCl _ (s) produced ? (at. mass K =39, Cl =35 5 ~g ~mol ⁻¹ )

Step-by-step solution

aligned 2 KClO _ 3(s) & 2 KCl _ (s) + 3 O _ 2(s) & 2 74.5 ~g 3 22.4 ~L &=149 ~g =67.2 ~L aligned Now, 67.2 ~L of O ₂=149 ~g of KCl at STP 33.6 ~L of O ₂= x g of KCl x= 149 33.6 67.2 =74.5 ~g of KCl