Ammonia gas is synthesized according to the balanced equation below. N2 (g) + 3 H2 (g) → 2 NH3 (g) If 2.75 L N2 react with 7.75 L H2, what is the theoretical yield (in liters) of NH3?

Volumes of reactants and products are assumed to be measured at the same temperature and pressure. (Should I convert the liters into grams to determine theoretical yield or should I leave them as they are?)

1 Answer
Jan 16, 2017

The balanced equation of the gaseous reaction given is

${N}_{2} \left(g\right) + 3 {H}_{2} \left(g\right) \to 2 N {H}_{3} \left(g\right)$

As per the reaction stochiometry we see that the gaseous ractants and products are in the following mole ratio $1 : 3 : 2$ So ratio of their volume will also follow same ratio.

Hence to maintain that ratio the $2.75 L \text{ } {N}_{2} \left(g\right)$ will react with $8.25 L \text{ } {H}_{2} \left(g\right)$ to form theoretically $2 \cdot 2.75 L = 5.50 L \text{ } N {H}_{3} \left(g\right)$
But the volume of ${H}_{2} \left(g\right)$ will fall short then.So hydrogen given will be fully consumed and the yeild of $N {H}_{3} \left(g\right)$ to be calculated on the consumption of of available ${H}_{2} \left(g\right) i . e .7 .75 L$

Si theoretical yeild of $N {H}_{3} \left(g\right)$ will be $= \frac{2}{3} \times 7.75 L \approx 5.17 L$