If 5.10 g of sodium and 305 g of potassium nitrate react in an airbag, how many grams of kno3 remain because of the limited amount of sodium present? 10na(s) 2kno3(s) k2o(s) 5na2o(s) n2(g)

The balanced equation for the above reaction is as follows;
10Na(s) +  2KNO₃(s)-->  K₂O(s) +  5Na₂O(s) +  N₂(g)
Stoichiometry of Na to KNO₃ 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

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