Question

Be sure to answer all parts. What is the mass of the solid NH4Cl formed when 76.5 g of NH3 is mixed with an equal mass of HCl? What is the volume of the gas remaining, measured at 14.0°C and 752 mmHg? What gas is it? NH3(g) + HCl(g) → NH4Cl(s) What is the mass of the NH4Cl produced?

Which gas remains? HCl or NH3

What is the volume of the gas remaining?

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)= 76.5 g

use:

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(76.5 g)/(17.03 g/mol)

= 4.491 mol

Molar mass of HCl,

MM = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass(HCl)= 76.5 g

use:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(76.5 g)/(36.46 g/mol)

= 2.098 mol

Balanced chemical equation is:

NH3 + HCl ---> NH4Cl +

1 mol of NH3 reacts with 1 mol of HCl

for 4.491 mol of NH3, 4.491 mol of HCl is required

But we have 2.098 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)*2.098

= 2.098 mol

use:

mass of NH4Cl = number of mol * molar mass

= 2.098*53.49

= 1.122*10^2 g

Answer: 112 g

b)

HCl is limiting reagent

So, NH3 would be remaining

Answer: NH3

c)

According to balanced equation

mol of NH3 reacted = (1/1)* moles of HCl

= (1/1)*2.098

= 2.098 mol

mol of NH3 remaining = mol initially present - mol reacted

mol of NH3 remaining = 4.491 - 2.098

mol of NH3 remaining = 2.393 mol

Given:

P = 752.0 mm Hg

= (752.0/760) atm

= 0.9895 atm

n = 2.393 mol

T = 14.0 oC

= (14.0+273) K

= 287 K

use:

P * V = n*R*T

0.9895 atm * V = 2.393 mol* 0.08206 atm.L/mol.K * 287 K

V = 56.9576 L

Answer: 57.0 L