Gases react on the mole level in these problems. If you are given the mass of th

Question

Gases react on the mole level in these problems. If you are given the mass of the gases, see the first example. In these problems we are interested in the variables that control the behavior of gases: P,V, n, T. You will use the ideal gas equation to solve for moles here. The problem reads: A 15.0 mL sample of ammonia gas at 100 torr and 30°C is mixed with 25 mL. of hydrogen chloride at 150. torr and 25°C and the reaction a) NH,a) +HCIw NH.CL. Calculate the mass of ammonium chloride that forms. Identify the gas in excess and determine its pressure (in the combined volume of the original two flasks) at 270C. Broken down into parts: (1) Convert pressures to atmospheres (2) Convert mers to liters (3) Convert temperatures from Celsius to kelvin (4) Use the ideal gas law to find the individual moles of reactants (5) Calculate the molar mass of ammonium chloride

Explanation / Answer

The reacion is NH3 + HCl ------------> NH4Cl

a) moles of NH3

V = 15mL = 0.015L

P = 100torr = 100/760 atm

T = 30C = 30+273=303 K

Ideal gas equation is PV = nRT

thus n=moles of NH3 = PV/Rt

=( 100/760)atmX0.015 L/ 0.0821L.atm/K.mol x303K

=7.93x 10-5 mol

b) moles of HCl

P= 150 torr = 150/760 atm

V = 25 mL = 0.025 L

T = 25 C = 25+273 = 298 K

thus n = PV/RT using ideal gas equation.

moles of HCl n = (150/760) atm x 0.025L /0.0821L.atm/K.mol x 298K

=2.01x10-4 mol

c) The moles of Nh3 are less than moles of HCl

Thus NH3 is the limiting reagent and the moles of NH4 Cl formed = moles of NH3

=7.93x 10-5 mol

molar mass of NH4Cl = 53.5 g/mol

thus mass of NH4Cl formed = moles formed x molar mass

= 7.93x 10-5 mol x53.5g/mol

= 4.24x10-3 g

Excess reagent is HCl = 2.01x10-4 mol -7.93x 10-5 mol

= 2.01x10-4 mol -0.793x10-4 mol

= 1.307x10-4 mol

total volume = 15+25 = 40mL = 0.040L

T = 27C = 27 +273 = 300 K

thus P = nRT/V

= [1.307x10-4 mol x0.0821L.at/k.mol x300K ] / 0.040L

=0.0804 atm

= 61.16 torr

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