Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. If 18.1 g of Nh3 is r

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Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. If 18.1 g of Nh3 is r

eacted with 90.4 g of CuO, which is the limiting reactant? How many grams of N2 will be formed? Explain how you solved for your answers.
Balanced Equation: 2NH3 + 3CuO → 3Cu + N2 + 3H2O

Chemistry

1 answer:

Scilla [17] · Answer 1 · 2023-10-06T09:32:52+03:00

I first converted the given grams of the reactants into moles, and then divided the moles by the coefficients in front of each of the reactant. The result with the smallest value will be the limiting reactant, and the value of CuO was the smallest, so it's the limiting reactant.

After figuring out which reactant is the limiting one, I took their given grams and converted it into moles, the divided it by the ratio of N2 to CuO (it's in the equation) to obtain the moles of N2, and then multiply it with the molar mass of N2 to get its mass in grams.