Unit 6 Test Review

Question

Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. If a sample of 18.1g of ammonia is reacted with 90.4g of copper (II) oxide, how much nitrogen gas will form?

NH₃ + CuO → N₂ + Cu + H₂O (unbalanced)



18.1

 

90.4

Which reactant is the limiting reactant? Enter a 1 for ammonia or a 2 for copper (II) oxide. ____

How many gram of N₂ are actually produced? ____g 

Answer 1

Hello, Sam,

There are several steps in this problem. I'm wondering which part is causing the most difficulty. If you've balanced the equation and calculated available moles of the reactants, please post those results with your question. Otherwise, it takes too much space (and time) to detail every step. Thanks.

I get a balanced equation of:

2NH₃ + 3CuO → N₂ + 3Cu + 3H₂O

There are 1.06 moles of NH₃ and 1.13 moles of CuO. The equation says we need 3 moles of CuO for every 2 moles of ammonia:

Molar ratio = (3 moles CuO)/(2 moles NH₃)

There isn't enough CuO to react will all the NH₃, so CuO is the limiting reagent.

We will assume all of the NH₃ is consumed, The molar ratio of N₂ to NH₃ is:

(1 mole N₂)/(2 moles NH₃)

Since we started with 1.o6 moles of NH₃, we should expect (1/2) that of N₂, or 0.53 moles N₂.

Convert to grams by multiplying by nitrogen's molar mass of 28 g/mole. We would predict 14.9 grams of nitrogen.

Bob