For the reaction,$2Fe(NO_3)_3 + 3Na_2CO_3 \to Fe_2(CO_3)_3 + 6NaNO_3$. Initially,if $2.5 \ mol$ of $Fe(NO_3)_3$ and $3.6 \ mol$ of $Na_2CO_3$ are taken. If $6.3 \ mol$ of $NaNO_3$ is obtained,then the $\%$ yield of the given reaction is:
- A$50$
- B$84$
- C$87.5$
- D$100$
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