2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l) a. What is the limiting reagent when 9.05 g of NH3 reacted with 45.2 g of CuO?(5 points) b. How many grams of N2 can be made?(10 points) c. If 4.61 g of N2 are made, what is the percent yield? (5 points) d. What is the mass of the excess reactant that remains after the reaction. (10 points)

a. Copper (II) oxide is the limiting reactant.

b.

c.

d.

Explanation:

Hello,

In this case, for the given reaction:

a. The limiting reactant is identified by computing the available moles of ammonia and the moles of ammonia that react with 45.2 g of copper (I) oxide as shown below:

In such a way, as there more ammonia available than that is reacted, we say it is in excess and the copper (II) oxide the limiting reactant.

b. Here, with the reacting moles of ammonia, we compute the yielded grams of nitrogen:

c. Now, since the 5.30 g of nitrogen are the expected grams of it, the percent yield of nitrogen is compute by dividing the real obtained mass over the theoretical previously computed mass:

d. Finally, as 0.532 moles of ammonia are available, but just 0.379 moles react, the unreacted moles are:

That in grams are:

Best regards.

answer; 4 electrons;