Question

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

A 5.49 g sample of NH3 gas and a 5.49 g sample of HCl gas are mixed in a 1.50 L flask at 25 degrees Celsius.

Identify the limiting reagent NH4Cl, NH3, HCl.

How many grams of will be formed by this reaction? mass: g

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

Answer 1

Sol :-

As,

Number of moles = Given mass in g / Gram molar mass

So,

Number of moles of NH₃ = 5.49 g / 17 g/mol

= 0.323 mol

Similarly,

Number of moles of HCl = 5.49 / 36.5 g/mol

= 0.150 mol

ICF table is :

..........................NH₃ (g)................. +......................... HCl (g) ------------------> NH₄Cl (s)

Initial (I)...............0.323 mol..........................................0.150 mol.........................0.0 mol

Change (C).........-0.150 mol............................................-0.150 mol.....................+0.150 mol

Final (F)...............0.173 mol..............................................0.0 mol..........................0.150 mol

Therefore,

Limiting reactant (Which is completely consumed in the reaction ) = HCl (g)

Mass of NH₄Cl formed = Moles x Gram molar mass

= 0.150 mol x 53.491 g/mol

= 8.02365 g

Hence, Mass of NH₄Cl formed = 8.02365 g

Moles of NH₃ (g) left (n) = 0.173 mol

From ideal gas equation :

PV = nRT

P = nRT/V

P = (0.173 mol).(0.0821 L atm K^-1mol^-1).(273.15 + 25) K / (1.50 L)

P = 2.823 atm

Hence, Pressure of NH₃ gas left = 2.82 atm