The balanced equation for the above reaction is as follows; <span>10Na(s) + 2KNO</span>₃(s)--> K₂O(s) + 5Na₂O(s) + N₂<span>(g) Stoichiometry of Na to KNO</span>₃ is 10 : 2 Number of Na moles reacted - 5.1 g/ 23 g/mol = 0.22 mol Na is the limiting reactant, therefore Na is fully used up, Since KNO₃ is present in excess , at the end of the reaction a certain amount of KNO₃ will be remaining. 10 mol of Na reacts with 2 mol of KNO₃ Therefore 0.22 mol of Na reacts with - 2 /10 x 0.22 = 0.044 mol of KNO₃ Mass of KNO₃ reacted = 0.044 mol x 101.1 g/mol = 4.45 g Mass of KNO₃ present initially - 305 g Therefore remaining mass of KNO₃ - 305 - 4.45 = 300.55 g
