the balanced equation for the above reaction is as follows;
2NaClO₃ ---> 2NaCl + 3O₂
stoichiometry of NaClO₃ to O₂ is 2:3
number of moles of NaClO₃ - 12.5 g / 106.4 g/mol = 0.11748 mol
according to molar ratio
if 2 mol of NaClO₃ gives 3 mol of O₂
then 0.117 mol of NaClO₃ will form - 3/2 x 0.11748 = 0.17622 mol
mass of O₂ formed - 0.176 mol x 32 g/mol = 5.64 g
mass of O₂ formed is 5.64 g
