If 38.7 grams of lithium chlorate decomposes, how many grams of oxygen gas can be produced?

The formula for lithium chlorate is:    2LiClO₃   →   2LiCl   +   3O₂

If moles =  mass ÷ molar mass  

then moles of LiClO₃  =  38.7 g  ÷  [( 7 × 1 ) + ( 35.5 × 1) + ( 16 × 3)] g/mol
                     
                                   =  38.7 g  ÷  ( 90.5 g ) g/mol

                                   =  0.428 mol

Now the mole ratio of   LiClO₃   :   O₂     is    2   :   3

∴  if moles of LiClO₃  = 0.428 mol

  then moles of O₂     =  ( 0.428 mol ÷ 2 ) × 3
          
                                  = 0.642 mol

Now since mass = moles × molar mass

∴ the mass of oxygen produced  =  0.642 mol × ( 16 × 2) g/mol

                                                    =  20.544 g

∴ the grams of oxygen gas produced when 38.7 grams of lithium chlorate is decomposed is ~ 20.54 g.

Most likely a is the answer - because  you generate   38 atp's but also consume an atp.   each atp gives approx. less than 4 kcal so probably a is the closest answer.