For the reaction , `2Fe(NO_(3))_(3)+3Na_(2)CO_(3) to Fe_(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 obatined then `%` yield of given reaction is:

Options

A. 50
B. 84
C. 87.5
D. 100

Answer

C

Explanation

`{:(,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`