Question

Ammonia and hydrogen chloride react to form solid ammonium chloride, nh4cl(s). two 1.00l flasks at 25c are connected by a stopcock. one flask contains 12.2 atm of nh3, and the other contains 9.78 atm hcl. When the stopcock is opened, the gases react until one is completely consumed. What will the final pressure of the system be, in atm, after the reaction is complete? (neglect the volume of the ammonium chloride formed.)

Answer 1

NH3 HCl -- > NH4Cl (S)

Calculate the moles of NH3 nad HCl as follows:

Moles of NH3:

PV= nRT

n= PV/RT

Here P = 12.2 atm, T = 25c = 298 K, and R = 0.0821 L atm / K.mol and V= 1.00 L

n= 12.2*1.00/0.0821*298 moles

n=0.499 or 0.5 mol

Moles of HCl:

PV= nRT

n= PV/RT

n= 9.78*1.00/0.0821*298 moles

n=0.399 or 0.4mol

Thus here HCl is limiting agent.

Here P = 9.78 atm, T = 25c = 298 K, and R = 0.0821 L atm / K.mol and V= 1.00 L

moles NH3 unreacted = 0.5 mol – 0.4 mol= 0.1mol which creates pressure only because NH4Cl is solid.

Use the idea gas law to determine the final pressure

PV = nRT

n = moles gas in final system = 0.1 mol
V = total volume final system = 1.00 L + 1.00 L = 2.00 L
T = temp in Kelvin = 25 + 273= 298 K
R = gas constant = 0.082057 L atm mol^-1 K^-1
P = pressure = ? atm

PV= nRT

P = nRT/V

P = 0.1 *0.0821*298/ 2

P= 1.22 atm final pressure