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For the reaction, 2Fe(NO3)+3Na2CO3rarrFe...

For the reaction, `2Fe(NO_3)+3Na_2CO_3rarrFe_2(CO_3)_3+6NaNO_3` initially `2.5` mole of `Fe(NO_3)_3 and 3.6` mole of `Na_2 CO_3` are taken. If `6.3` mole of `NaNO_3` is obtained then % yield of given reaction is :

A

50

B

84

C

87.5

D

100

Text Solution

Verified by Experts

The correct Answer is:
C
`{:(,2Fe(NO_(3))_(2),+,3Na_(2)CO_(3),rarr,Fe_(2)(CO_(3))_(3),+,6NaNO_(3)),("mole",2.5,,3.5,,,,),("mole/stoichiometric coefficient",1.25,,1.2,,,,):}`
Limiting reagent is `Na_(2)CO_(3)` so moles of `NaNO_(3)` formed `= 3.6 xx 2 = 7.2` % yield `= (6.3)/(7.2) xx 100 = 87.5`
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