Solid ammonium chloride, NH 4 Cl , is formed by the reaction of gaseous ammonia, NH 3 , and hydrogen chloride, HCl . NH 3 ( g ) + HCl ( g ) ⟶ NH 4 Cl ( s )

Question

A 6.90 g sample of NH 3 gas and a 6.90 g sample of HCl gas are mixed in a 2.00 L flask at 25C.

Identify the limiting reagent.

NH4Cl

or

HCl

or

NH 3

How many grams of  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 this reaction.

Answer 1

Write the balanced equation:

NH3(g) + HCl(g) ==> NH4Cl(s)

Find limiting reactant:

6.90 g NH3 x 1 mol NH3/17 g x 1 mol NH4Cl/mol NH3 x 52.5 g/mol NH4Cl = 21.3 g NH4Cl formed

6.90 g HCl x 1 mol HCl/36.5 g x 1 mol NH4Cl/mol HCl x 52.5 g/mol NH4Cl = 9.92 g NH4Cl formed

HCl IS THE LIMITING REACTANT and 9.92 g of NH4Cl will be formed.

Moles NH3 gas remaining at end of reaction:

Starting moles NH3 = 6.90 g NH3 x 1 mol NH3/17 g = 0.406 moles NH3

6.90 g HCl x 1 mol HCl/36.5 g x 1 mol NH3/mol HCl = 0.189 moles NH3 used up to react with the HCl

moles NH3 remaining = 0.406 - 0.189 = 0.217 moles NH3 remaining

Using the ideal gas law to find the pressure of the gas remaining:

PV = nRT

P = nRT/V = (0.217)(0.0821 Latm/Kmol)(298K)/2.00 L

P = 2.65 atm