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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:

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a

50

b

84

c

87.5

d

100

answer is C.

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Detailed Solution

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

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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: