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 5.59 g sample of NH3 gas and a 5.59 sample of HCL gas are mixed in a 1.00 L flask at 25 degrees Celsius.

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

Hello, Emily,

The balanced equation tells us that 1 mole of both the NH₃ and HCl will produce 1 mole of NH₄Cl. Let's determine the actual number of moles of both reactants. We need their molar masses (g/mole). Divide the actual mass (g) by the molar mass to get moles: g/(g/mole) = moles

We don't have enough HCL (0.153 moles) to react with all of the NH₃ (0.329 moles). [3 sig figs]. Once the 0.153 moles of HCl is consumed, the reaction will stop. Since the molar ratios are 1:1:1, we'll consume only 0.153 moles of the NH₃, leaving the rest unreacted. ).153 moles of NH₄Cl are produced. To convert this to grams HCl, multiply the moles HCl by it's molar mass (0.153 moles HCl)*(53.5 g HCl/mole HCL)= 8.19 grams HCl (3 sig figs).

The pressure calculation is a bit trickier. I'm not certain what is meant by the "gas remaining in the flask." Does this mean only the unreacted NH₃? I will assume there is no air in the flask.

Moles NH₃ = (0.329 - 0.153)= 0.153 moles of NH₃ that are unreacted.

We ignore the NH₄Cl solid.

The gas law:

PV = nRT, where n is moles, T is temperature in Kelvin, R is the gas constant, and P is the pressure.

P = (0.153 moles)*(0.176 moles NH3)*(0.08205 L*atm/mole*K)*(298K)

P = 4.30 atmospheres

This seems high to me. So I've either entered a wrong value, or I hope the flask is made of 1" thick stainless steel.

I hope some part of this helps,

Bob