Question

Hydrogen gas is produced by the reaction between metallic aluminum and aqueous hydrochloric acid. 2 Al(s) + 6 HCl(aq) — 2 AICI, (aq) + 3 H, (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 294.9 mL of gas with a total pressure of 1.33 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. A How many grams of aluminum must have reacted to produce this quantity of hydrogen gas? mass of aluminum:

Answer 1

total pressure (PT) = 1.33atm

The vapor pressure of water at 29.4⁰C (PH2O) = 30.75 torr

                                                                               = 30.75/760   = 0.04atm

The partial pressure of H2 ( PH2) = PT-PH2O

                                                      = 1.33-0.04

                                                       = 1.29atm

PH2 = 1.29atm

V   = 294.9ml = 0.2949L

T   = 29.4+273   = 302.4K

PV   = nRT

n   = PV/RT

        = 1.29*0.2949/(0.0821*302.4)

         = 0.01532moles

2Al(s) + 6HCl(aq) -------------> 2AlCl3(aq) + 3H2(g)

3 moles of H2 produced from 2 moles of Al react with HCl

0.01532 moles of H2 produced from = 2*0.01532/3   = 0.0102 moles of Al react with HCl

mass of Al = no of moles * gram molar mass

                    = 0.0102*27   = 0.2754g

mass of aluminum   = 0.2754g >>>>answer