Question

Ammonia combines with acids in the presence of water to produce salt.
NH3(g) + HCI(aq) ----> NH4CI(aq)
If 45 grams of NH3 are used with an excess of HCI, how much grams of ammonia chloride will be produced?
A.45g
B.33.8g
C.141.5g
D.2.65g

Answer 1

First of all you have to find amounts in mol of all known components of reaction:

HCl - excess

NH₃ 45 g

n of NH₃ = m/M_w = 45 / 17 = 2.647 mol

NH₃(g) + HCI(aq) ----> NH₄CI(aq)

the ratio between NH₃ : NH₄CI = 1:1, so amount
of NH₄CI is 2.647 mol

Now you can find mass of NH₄CI

m = n*M_w = 2.647*53.5 = 141.5g

So the answer is C = 141.5g