Solid ammonium chloride, NH4Cl, is formed by the reaction of gaseous ammonia, NH3, and hydrogen chloride, HCl.

Question

NH3(g)+HCl(g)⟶NH4Cl(s)

A 6.44 g sample of NH3 gas and a 6.44 g sample of HCl gas are mixed in a 0.50 L flask at 25 ∘C.

Identify the limiting reagent.

How many grams of NH4Cl will be formed by this reaction?

What is the pressure in atmospheres of the gas remaining in the flask? Ignore the volume of solid NH4Cl produced by the reaction.

Answer 1

NH₃(g) + HCl(g) ==> NH₄Cl(s) ... balanced equation

One easy way to identify the limiting reagent is to divided the moles of each reactant by its coefficient in the balanced equation and whichever value is less is the limiting reagent. In the current problem we have...

NH₃: 6.44 g NH₃ x 1 mol NH₃/17 g = 0.3688 moles NH₃ (÷ 1 = 0.3688)

HCl: 6.44 g HCl x 1 mol HCl/36.5 g = 0.1764 moles HCl (÷ 1 = 0.1764) -> LIMITING REAGENT

How many grams of NH₄Cl will be formed by this reaction?

0.1764 moles HCl x 1 mole NH₄Cl / mole HCl x 53.49 g NH₄Cl/mole = 9.44 g NH₄Cl (3 sig. figs.)

What is the pressure in atmospheres of the gas remaining in the flask?

NH₃(g) + HCl(g) ==> NH₄Cl(s)

0.3688......0.1764............0..........Initial

-0.1764....-0.1764........+0.1764...Change

0.1924 .......0...............+0.1924....Equilibrium

The only gas remaining in the flask is NH₃ and there are 0.1924 moles. This is in a volume of 0.5 L at a temperature of 25ºC (+273 = 298K). Using the ideal gas law, we can then solve for the pressure (P).

PV = nRT

P = nRT/V = (0.1924 mol)(0.0821 Latm/Kmol)(298K) / 0.5 L

P = 9.41 atm