How many grams of oxygen are formed when 58.6 g of kno3 decomposes?

The formula for potassium nitrate is:    2KNO₃   →   2KNO₂   +   O₂

If moles =  mass ÷ molar mass  

then moles of KNO₃  =  58.6 g  ÷  [( 39 × 1 ) + ( 14 × 1) + ( 16 × 3)] g/mol
                     
                                   =  58.6 g  ÷  ( 101 g ) g/mol

                                   =  0.5802 mol

Now the mole ratio of   KNO₃   :   O₂     is    2   :   1

∴  if moles of  KNO₃ = 0.5802 mol

  then moles of O₂     =  0.5802  mol ÷ 2 
          
                                  = 0.2901 mol

Now since mass = moles × molar mass

∴ the mass of oxygen produced  =  0.2901 mol × ( 16 × 2) g/mol

                                                    = 9.283 g

∴ the grams of oxygen gas produced when 58.6 grams of potassium nitrate is decomposed is ~ 9.28 g