Question

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

2Al(s)+6HCl(aq)⟶2AlCl3(aq)+3H2(g)$$\mathrm{}\mathrm{}\mathrm{}{}_{\mathrm{}}^{}\mathrm{}{}_{\mathrm{}}^{}$$

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 292.0$\mathrm{}$ mL of gas with a total pressure of 1.33$\mathrm{}$ atm is collected via water displacement at 29.4 ∘C$\mathrm{}{}^{}$, what is the partial pressure of the hydrogen gas in the sample? The vapor pressure of water at 29.4 ∘C$\mathrm{}{}^{}$ is 30.75$\mathrm{}$ torr.

Answer 1

Given: P_water = 30.75torr @ 29.4^oC ; P_total = 1.33 atm

Total pressure of the displaced gas is the partial pressure of H₂ and the vapor pressure. 

P_total = P_water + P_H2

Step 1: Convert P_water from  torr to atm: 

1.00atm = 760torr

P_water , atm = (30.75torr)(1.00atm/760torr)

P_water , atm = 0.04046 atm

Step 2: Solve for P_H2 

P_H2 = P_total - P_water 

P_H2 = 1.33 atm - 0.04046 atm

P_H2 = 1.28₉ atm

P_H2 = 1.29 atm

We are only asked to solve for the partial pressure of Hydrogen gas in the displaced gas. So we do not need to use the ideal gas equation to solve it because in the first place the moles of the gas was not given. 

With the partial pressure of H₂ we can solve for the moles of the H₂ produced and the amounts of the reactants given that the amount of H₂ is in a hundred percent yield.