Ammonia and hydrogen chloride react to form solid ammonium chloride: NH_3(g) + H

Question

Ammonia and hydrogen chloride react to form solid ammonium chloride: NH_3(g) + HCl(g) rightarrow NH_4Cl(s) Two 2.00-L flasks at 25 degree C are connected by a valve. One flask contains 5.00 g of NH_3(g) and the other contains 5.00 g of HCl(g). When the valve is opened, the gases react until one is completely consumed. Which gas will remain in the system after the reaction is complete? (b) What will be the final pressure of the system after the reaction is complete? (Neglect the volume of the ammonium chloride formed.)

Explanation / Answer

Total volume of two flasks together = 4L

Temperature =298 K

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

5/17 5/36.5 g 0 initial moles

Since 1 mole of NH3 reacts with 1 mole of HCl

a) In the given reaction, the amount of HCl(0.136Moles) is less than required(0.29) .Hence Hcl is the limiting reagent.and is consumed completely . Ammonia is remained unreacted.

b)

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

5/17 5/36.5 g 0 initial moles

= 0.295 = 0.136

0.159 0 0.136 equilibrium moles

Thus we are left with 0.159 mole of NH3 in 4L volume at 298K

We ccan calulate the Pressure of the system by PV = nRT

Therefore P = nRT/V = 0.159 xo.o821 x298 /4

= 0.972 atm

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