For the reaction 2FeNO33+3Na2CO3⟶Fe2CO33+6NaNO3 Initially if 2.5 mole of Fe(NO3)3 and 3.6 mole of Na2CO3 is taken. If 6.3 mole of NaNO3 is obtained then % yield of given reaction is:

mole 2FeNO332.5+3Na2CO33.6⟶Fe2CO33+6NaNO32.5 moles of Fe(NO3)3 require 3.75moles of Na2CO3Hence Limiting reagent is Na2CO3 so moles of NaNO3 should be formed =3.6 x 2 =7.2%yield =6.37.2×100=87.5