Ammonia and hydrogen chloride react to form solid ammonium chloride: NH3(g) + HC

Question

Ammonia and hydrogen chloride react to form solid ammonium chloride:

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

Two 2.00-L flask at 25 degrees celcius are connected by a valve. One flask contains 5.00 g of NH3(g) and the other contains 5.00 g of HCl(g). When the valve is opened, the gases react until one is completly consumed.

a) Which gas will remain in the system after the reaction is complete?

b) What will be the final pressure of the systerm after the reaction is complete? (neglect the volume of the solid ammonium chloride formed)

Explanation / Answer

Calculate moles of NH3 and HCl

Mol = mass in g /molar mass

molar mass of NH3 = 17.031 g /mol

Molar mas of HCl = 36.46094 g/mol

n HCl = 5.0 g HCl x 1 mol HCl / 36.4609 g HCl

5.0/36.4609=0.1371 mol HCl

n NH3 = 5.00/17.031=0.2936 mol NH3

In the given reaction, mole ratio is 1:1 so mol ratio between NH3:HCl should be 1 but its not true if we look at given moles.

Moles of HCl is less than moles of NH3 so HCl is limiting reactant and moles of NH3 will remain in the container even after the reaction.

a) NH3 will remain after the reaction is completed.

b) For this part, we need to calculate moles of NH3 in excess

Excess moles of NH3 = original moles present - moles of NH3 required to react with HCl

= 0.2936-0.1371=0.1565 mol NH3

Les use pV=nRT   (ideal gas law)

Here p is pressure , V is volume in L, n is moles , R is gas constant, T is temperature in K.

R= 0.08206 L atm/K mol

Lets plug in the values

   p = (0.1565*0.08206*298.15)/2.00=1.91448 atm

Pressure = 1.9 atm

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