Question

In aqueous solution HCL and Al react to produce H2 and AlCl3. A 5g impure sample of Al is dissolved completely in water then treated with 2 mol L-1 HCL to evolve 6409 ML of H2 at 25C and 100kpa. The HCL solution is slowly added until the evolution of H2 is no longer observed. What percentage (by mass) of the impure sample is Al?

Answer 1

moles of H2 generated= PV/RT, P= 100Kpa= 100/101.3 atm (101.3 Kpa= 1atm)= 0.987

V= 6409ml= 6409/1000L =6.409L (1000ml=1L), T= 25 deg.c= 25+273=298K, R=0.0821 L.atm/mole.K(gas constant)

n= 0.987*6.409/(0.0821*298)= 0.26 moles

the reaction for producing H2 is 2Al+6HCl----->2AlCl3+3H2

since there is enough HCl to ensure all the Al reacts,

the reaction suggests 3 moles of H2 is generated from 2 moles of Al

0.26 moles of H2 requires 0.26*2/3 moles of Al =0.173

mass of Al in the sample= moles of Al* atomic weight of Al =0.173*27= 4.671gm

mass % of Al in the sampple= 100*(mass of Al in the sample/ total mass)=100*4.671/5=93.42%