Question

Ammonia, NH₃, is produced from the reaction of gaseous nitrogen and hydrogen at high temperatures.

3 H_{2 (g)} + N_{2 (g)}  → 2 NH_{3 (g)}

For the reaction of 3.77 g of H₂ with 8.66 g of N₂, you determined that 21.2 g and 10.5 g of ammonia could be produced, respectively. In the laboratory, you reacted these masses of H₂ and N₂ and collected 7.70 g of NH₃. What is the percentage yield of this reaction to the correct number of significant figures?

_____Units_______

Answer 1

3 H_{2 (g)} + N_{2 (g)}  → 2 NH_{3 (g)}

No. of moles of H2 = 3.77 / 2 = 1.885 moles

No. of moles of N2 = 8.66 / 28 = 0.309 moles

no. of moles of H2 needed to react with 0.309 moles of N2 = 3 x 0.309 = 0.928 moles

Here, No. of moles of H2 present in excess.

So, limiting reagent is N2

No. of moles of NH3 produced = 2 x 0.309 = 0.618 moles

Weight of NH3 produced = 0.618 x 17 = 10.515 grams

Actual weight of NH3 collected = 7.70 g

Percentage yield = (7.70 / 10.515) x 100 = 73.2 %