Question

Solid ammonium chloride, NH4Cl, is formed by the reaction of gaseous ammonia, NH3, and hydrogen chloride, HCl.NH3(g)+HCl(g)⟶NH4Cl(s)

A 6.37 g sample of NH3 gas and a 6.37 g sample of HCl gas are mixed in a 1.00 L flask at 25 ∘C.

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

1)

Molar mass of NH3,
MM = 1*MM(N) + 3*MM(H)
= 1*14.01 + 3*1.008
= 17.034 g/mol


mass(NH3)= 6.37 g

use:
number of mol of NH3,
n = mass of NH3/molar mass of NH3
=(6.37 g)/(17.03 g/mol)
= 0.374 mol

Molar mass of HCl,
MM = 1*MM(H) + 1*MM(Cl)
= 1*1.008 + 1*35.45
= 36.458 g/mol


mass(HCl)= 6.37 g

use:
number of mol of HCl,
n = mass of HCl/molar mass of HCl
=(6.37 g)/(36.46 g/mol)
= 0.1747 mol
Balanced chemical equation is:
NH3 + HCl ---> NH4Cl +


1 mol of NH3 reacts with 1 mol of HCl
for 0.374 mol of NH3, 0.374 mol of HCl is required
But we have 0.1747 mol of HCl

so, HCl is limiting reagent
we will use HCl in further calculation


Molar mass of NH4Cl,
MM = 1*MM(N) + 4*MM(H) + 1*MM(Cl)
= 1*14.01 + 4*1.008 + 1*35.45
= 53.492 g/mol

According to balanced equation
mol of NH4Cl formed = (1/1)* moles of HCl
= (1/1)*0.1747
= 0.1747 mol


use:
mass of NH4Cl = number of mol * molar mass
= 0.1747*53.49
= 9.346 g
Answer: 9.35 g

2)
According to balanced equation
mol of NH3 reacted = (1/1)* moles of HCl
= (1/1)*0.1747
= 0.1747 mol
mol of NH3 remaining = mol initially present - mol reacted
mol of NH3 remaining = 0.374 - 0.1747
mol of NH3 remaining = 0.1992 mol

Given:
V = 1.0 L
n = 0.1992 mol
T = 25.0 oC
= (25.0+273) K
= 298 K

use:
P * V = n*R*T
P * 1 L = 0.1992 mol* 0.08206 atm.L/mol.K * 298 K
P = 4.87 atm
Answer: 4.87 atm