Ammonia, NH3(g), and hydrogen chloride, HCl(g), react to form solid ammonium chl

Question

Ammonia, NH3(g), and hydrogen chloride, HCl(g), react to form solid ammonium chloride, NH4Cl(s): NH3(g)+HCl(g)NH4Cl(s) Two 3.00 L flasks at 35.0 C are connected by a stopcock, as shown in the drawing. One flask contains 5.40gNH3(g), and the other contains 5.00 g HCl(g). When the stopcock is opened, the gases react until one is completely consumed. 1.What will be the final pressure of the system after the reaction is complete? (Neglect the volume of the ammonium chloride formed.) 2.What mass of ammonium chloride will be formed?

Explanation / Answer

Given

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

Given

Volume V1 = V2 = 3 L

Temperature = 35 C

Mass of NH3 = 5.4 g

Molar mass of NH3 = 17 g/mol

No. of moles = Mass / Molar mass = 5.4 g / 17 g/mol = 0.318 moles

Mass of HCl = 5.4 g

Molar mass of HCl = 36.5 g/mol

No. of moles = Mass / Molar mass = 5. g / 36.5 g/mol = 0.137 moles

from the given reaction each mole of NH3 requires 1 mole of HCl

so 0.137 moles of HCl will require 0.137 moles of NH3

remaining 0.318 - 0.137 = 0.181 moles will remain as gas

only this will be responsible for the pressure in the total two flasks as all others are in solid form

No. of moles = 0.181 moles

Volume = 3 L + 3 L = 6 L

Temperature = 35 C = 308 K

PV = nRT

P * 6 L = 0.181 moles * 0.08206 L.atm/mol.K * 308 K

P = 0.762 atm Answer 1

0.137 moles of NH4Cl will be formed

Molar mass of NH4Cl = 53.5 g/mol

Mass = No. of moles * Molar mass = 0.137 moles * 53.5 g/mol = 7.33 g Answer 2

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