The reaction of metallic aluminum with aqueous hydrochlotic acid produces hydrog

Question

The reaction of metallic aluminum with aqueous hydrochlotic acid produces hydrogen gas.

2Al + 6HCl --> 2AlCl3 + 3H2

Hydrogen gas produced by this reaction is typically collected via water displacement, during which time the hydrogen gas becomes saturated with water vapor. A 241.5 mL sample of gas with a total pressure 118 kPa was collected via water displacement at 29.4C. Calculate the partial pressure of hydrogen gas in the sample. The vapor pressue of water at 29.4C is 4.10 kPa.

Calculate the mass of aluminum that reacted to produce this quantity of hydrogen gas.

Explanation / Answer

Total pressure of sample gas (H2 saturated with water vapor ) = 118 kPa
The vapor pressure of water vapor = 4.10 kPa
Therefore, the partial pressure of H2 gas in the sample = 118 - 4.10 kPa
   = 113.9 kPa
Now the Moles of H2 gas collected = PV/ RT
   = (113.9 kPax 0.2415 L)/ ((8.314 L-kPa/ mol-K)x(302.4 K)
= 0.011 mole
From the balanced reaction 2 mole Al reacts to form 3 moles of H2.
Therefore, to produce 0.011 mole H2 moles of Al reacted = (2/3) x 0.011
   = 7.33 x 10-3 moles Al
Molar mass of Al = 26.98 g/mol

Mass of Aluminium reacted = 26.98 g/mol x 7.33 x 10-3 mole
   = 0.1978 grams Al

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