Question

Hydrogen gas is produced by the reaction between metallic aluminum and aqueous hydrochloric acid.

2Al(s)+6HCl(aq)⟶2AlCl3(aq)+3H2(g)

The hydrogen gas produced by this reaction is typically collected via water displacement. During this process, the hydrogen gas becomes saturated with water vapor. If 281.7 mL of gas with a total pressure of 1.02 atm is collected via water displacement at 29.4 ∘C, what is the partial pressure of the hydrogen gas in the sample? The vapor pressure of water at 29.4 ∘C is 30.75 torr.

Find:

PH2= atm

How many grams of aluminum must have reacted to produce this quantity of hydrogen gas?

mass of aluminum:

Answer 1

According to Dalton's law of partial pressures

Total pressure = vapor pressure of water vapor + partial pressure of hydrogen gas

Mass of aluminium reacted = moles × molar mass = 7.41 × 10^-3 mol × 27 g/mol = 0.200 grams.

• Partial pressure of water vapors = 30.75 tory x atm 760 torr = 0.0405 atm atm + b (H₂) so, Ptotal = 1.02 atm = 0.0405 * Partial pressure of H₂ gas = 10.98 atml · Volume of He gas (V) = 281.7 ml x10 SL ml = 0.2817 L · Temperature (T) = 29.4°C = (29.4 + 273.15) k = 302:55 K i using Ideal gas equation, pV = nRT Moles of Hz gas, n = ÞV/RT 0.98 atm x 0. 2817 ml . 0.0821 L atm x 302.55 K K mol = 0.01 mol • from → the chemical equation, 3 mol H₂ ² 2 mol Al 0.0111 mol Hy 0.0111 mol Al = 7.41*10-3 mol