The mass of $\mathrm{N}_{2} \mathrm{~F}_{4}$ produced by the reaction of $2.0 \mathrm{~g}$ of…
The mass of $\mathrm{N}_{2} \mathrm{~F}_{4}$ produced by the reaction of $2.0 \mathrm{~g}$ of $\mathrm{NH}_{3}$ and $8.0 \mathrm{~g}$ of $\mathrm{F}_{2}$ is $3.56 \mathrm{~g}$. What is the percent yield of the final product formed i.e. \(\mathrm{N}_2 \mathrm{~F}_4\) ? $2 \mathrm{NH}_{3}+5 \mathrm{~F}_{2} \longrightarrow \mathrm{N}_{2} \mathrm{~F}_{4}+6 \mathrm{HF}$
$79.0$ %
$71.2$ %
$84.6$ %
None of these
Solution
$\underset{34 \mathrm{~g}}{2 \mathrm{NH}}(\mathrm{g})+\underset{190 \mathrm{~g}}{5 \mathrm{~F}_{2}} ightarrow \underset{104 \mathrm{~g}}{\mathrm{~N}_{2} \mathrm{~F}_{4}}+6 \mathrm{HF}$
Amount of $\mathrm{N}_{2} \mathrm{~F}_{4}$ formed by $2 \mathrm{~g} \mathrm{NH}_{3}=\frac{104}{34} \times 2=6.12$
Amount of $\mathrm{N}_{2} \mathrm{~F}_{4}$ formed by $8 \mathrm{~g} \mathrm{~F}_{2}=\frac{104}{190} \times 8=4.38$
$\mathrm{N}_{2} \mathrm{~F}_{4}$ will be limiting and actual amount of product is $3.56 \mathrm{~g}$
$\therefore \% \text { yield }=\frac{\text { Actual amount of product }}{\text { Calculated amount of product }} \times 100$
$=\frac{3.56}{4.38} \times 100=81.2896 ~\%$