1. The problem statement, all variables and given/known data 2 KClO3 (s) + heat --> 2 KCl (s) + 3 O2 (g) If 0.20 g of KClO3 (s) decomposes in 2.8 s, at what rate is oxygen O2 (g) released over this time at SATP? 3. The attempt at a solution 0.20 g KClO3 x 1 mol/122.5495 g = 0.001632 mol KClO3 Since 0.001632 mol decomposes in 2.8 seconds I found that 2 mol of it decomposes in 3431.37 seconds. So then to find the rate of decomposition of KClO3 I did 2/3431.37 = 0.000583 mol/s Now that I know the rate of decomposition of KClO3 I can find the rate of production of oxygen gas from the balanced reaction equation. 0.000583 mol/s - 2 mol n - 3 mol n = 0.000874 mol/s O2 (g) The answer in the back says 22 mL/s but I have know idea how that answer was gotten. Can someone please help?