Answer to Question #137039 in General Chemistry for nevar

Question

8. Based on the balanced equation, if 6.02 g of NH3 reacting with 22.3 g of CuO, which one is the limiting reagent? How many grams of N2 (g) can be made? If 2.51 g of N2 are actually made, what is the percent yield? (NH3=17.03, CuO=79.55, N2=28.02)2 NH3 (g) + 3 CuO (s) → N2 (g) + 3 Cu (s) + 3 H2O (l)

Answer 1

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

If 6.02 g of NH₃ are reacted with 22.3 g of CuO, which is the limiting reagent? How many grams of N₂ will be formed?

First we compute the number of moles of NH₃ (M.W. = 17.031 g/mole) and the number of moles of CuO (M.W. = 79.5 g/mole).

n(NH₃) = 6.02/17.031 = 0.35 mole

n(CuO) = 22.3/79.5 = 0.28 mole

To determine which reagent is limiting we use the mole ratio from the chemical equation to convert moles NH₃ to moles CuO.

0.35*3/2 = 0.525 mole CuO

Therefore CuO is the limiting reagent. That is, CuO will run out before the NH₃ does.

The mass of N₂ produced will be

n(N₂) = 0.28/3 = 0.09 mole

m(N₂) = 0.09*28 = 2.52 g

yield = 2.51/2.52 = 99.6%