Nitrogen gas can be prepared by passing gaseous ammonia over solid CuO at high t

Question

Nitrogen gas can be prepared by passing gaseous ammonia over solid CuO at high temperatures. The other products of the reation are solid copper and water vapor. If a sample containing 18.1g of NH3 is reacted with 90.4g of CuO, which is the limiting reactant? How many grams of N2 will be formed? What is the theoretical yield ? If 6.63g are actually producted what is the percent yield in this case?
I believe the balanced equation is:
2NH3(g) + 3CuO(s) ---> N2(g) + 3Cu(s) + 3H2O(g)

For moles NH3 I have: 18.1gNH3/17.03g/mol= 1.063 moles NH3

For moles CuO I have: 90.4gCuO/79.546g/mol= 1.14 moles CuO

Do I use the ratio between NH3 and CuO (2:3) to find the limiting reactant, or the ratio between 2NH3:N2 / 3CuO:N2?

Explanation / Answer

2NH3(g) + 3CuO(s) ---> N2(g) + 3Cu(s) + 3H2O(g)

from equation :

2 mole NH3 = 3 mole cUo = 1 mole N2 = 3 mole Cu = 3 mole H2o

no of moles NH3 = 18.1gNH3/17.03g/mol= 1.063 moles NH3.

no of moles CuO = 90.4gCuO/79.546g/mol = 1.14 moles CuO.

yes to determine the limiting reagent use 2:3 ratio

limiting reagent is Cuo

No of moles of N2 produced = 1.14/3 = 0.38 mole

mass of N2 = 0.38*28 = 10.64 grams

theoretical yield = 10.64 grams

practical yield = 6.63 grams

percentyield = practical yield/theoretical yield*100

    = 6.63/10.64*100 = 62.31%

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.