Question

For the reaction, $Fe(N{O}_3{)}_3+N{a}_{2}C{O}_3\to F{e}_2(C{O}_3{)}_3+NaN{O}_3$ Initially if 2.5 moles of $Fe(N{O}_3{)}_2$ and 3.6 moles of $N{a}_{2}C{O}_3$ is taken. If 6.3 moles of $NaN{O}_3$ is obtained then $\%$ yield of given reaction will be:

• A. $50\%$
• B. $84\%$
• C. $87.5\%$
• D. $100\%$

Answer 1

The correct option is C $87.5\%$

On balancing the given reaction is:
$2Fe(N{O}_3{)}_3+3N{a}_{2}C{O}_3\to F{e}_2(C{O}_3{)}_3+6NaN{O}_3$

Moles of $Fe(N{O}_3{)}_3$ = 2.5
Moles of $N{a}_{2}C{O}_3$ = 3.6

$\frac{\text{moles of}N{a}_{2}C{O}_3}{\text{stoichiometric coefficient}}=\frac{3.6}{3}=1.2$
$\frac{\text{moles of}Fe(N{O}_3{)}_3}{\text{stoichiometric coefficient}}=\frac{2.5}{2}=1.25$
Hence $N{a}_{2}C{O}_3$ is the limiting reagent.
$\therefore$ $\frac{molesofN{a}_{2}C{O}_3}{3}=\frac{molesofNaN{O}_3}{6}$


$\text{Moles of}NaN{O}_3\text{produced theoretically}=\frac{3.6}{3}\times 6=7.2mol$
Experimental yield = $6.3mol$
$\text{percentage yield}=\frac{6.3}{7.2}\times 100=87.5\%$