A sample of nitrogen gas was collected via water displacement. Since the nitroge

Question

A sample of nitrogen gas was collected via water displacement. Since the nitrogen was collected via water displacement, the sample is saturated with water vapor. If the total pressure of the mixture at 21 degree C is 1.47 atm, what is the partial pressure of nitrogen The vapor pressure of water at 21 degree C is 18.7 mm Hg. Hydrogen gas can produced by the reaction between metallic aluminum and aqueous hydrochloric acid 2Al(s)+6HCl(aq) right arrow 2 AlCl_3(aq) + 3H_2(g) Hydrogen gas produced by this reaction is typically collected via water displacement, during which time the hydrogen gas becomes saturated with water vapor. If 229.8 mL of gas with a total pressure 1.12 atm was collected via water displacement at 29.4 degree C, what is the partial pressure of hydrogen gas in the sample How many grams of aluminum must have reacted to produce this quantity of hydrogen gas The vapor pressure of water at 29.4 degree C is 30.75 torr.

Explanation / Answer

Solution :-

Q1). Total pressure = 1.47 atm

1.47 atm * 760 mmHg / 1 atm = 1117.2 mmHg

Vapore puressure of water = 18.7 mmHg

Pressure of the nitrogen = total pressure – vapor pressure of water

                                       = 1117.2 mmHg – 18.7 mmHg

                                      = 1098.5 mmHg

Lets convert mmHg to atm

1098.5 mmHg * 1 atm / 760 mmHg = 1.45 atm

So the pressure of the nitrogen is 1.45 atm

Q2). Total pressure = 1.12 atm

Temperature = 29.4 C

Vapor pressure of water at 29.4 C = 30 mmHg

30.75 mmHg * 1 atm / 760 mmHg = 0.04046 atm

Partial pressure of Hydrogen gas = total pressure – vapor pressure of water

                                                       = 1.12 atm – 0.04046 atm

                                                       = 1.08 atm

So the partial pressure of H2 gas = 1.08 atm

Now lets calculate the moles of the H2

PV= nRT

PV/RT= n

0.2298 L * 1.08 atm / 0.08206 L atm per mol K * 302.4 K = n

0.01 mol H2 = n

Now using the moles of H2 lets calculate moles of Al reacted

now lets convert moles of Al to its mass

mass of Al= moles * molar mass

                = 0.006667 mol * 26.982 g per mol

                = 0.180 g Al

So the mass of Al reacted is 0.180 g

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