For the reaction,2FeNO33+3Na2CO3→Fe2CO33+6NaNO3Initially 2.5 mole of Fe(No3)2 and 3.6 mole of Na2CO3 are taken. If 6.3 mole of NaNO3 is obtained then % yield of given reaction is:

2FeNO33+3Na2CO3→Fe2CO33+6NaNO3 mole  2.5            3.6 mole/  stoichiometric  coefficient 2.52=1.25 3.63=1.2Limiting reagent is Na2CO3So moles of NaNO3 formed = 3.6 x 2  = 7.2% yield =  Experimental yield ×100Calculated yield % yield = 6.37.2×100=87.5