If this reaction were run using 38 grams of NH3 and excess CuO, how many grams of water would you expect to form?

Answer 1

NH3 + CuO → N2 + Cu + H2O

Balance:

2NH3 + 3CuO → N2 + 3Cu + 3H2O

38 g NH3 x 1 mol NH3/17.03 g NH3 x 3 mol H2O/2 mol NH3 x 18.02 g H2O/1 mol H2O = ?

Solve this.

Answer 2

What reaction are you referring to? Perhaps the following:

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

moles of NH₃ used = 38 g NH₃ x 1 mole NH₃/17 g = 2.24 moles NH₃ used

Since CuO is in excess, the amount of H₂O formed will depend only on the moles of NH₃, i.e. 2.24 moles NH₃

moles H₂O formed = 2.24 moles NH₃ x 3 moles H₂O/2 moles NH₃ = 3.36 moles H₂O formed

mass H₂O formed = 3.36 moles H₂O x 18 g/mole = 60.48 g H₂O = 60. g H₂O formed (2 significant figures)