Iron-aluminum alloys are useful in some applications because they become magneti

Question

Iron-aluminum alloys are useful in some applications because they become magnetized in a magnetic field but are easily demagnetized when the field is removed. The composition of an iron-aluminum alloy can be determined chemically by reacting it with hydrocholoric acid:

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

Fe(s) + 2 HCl(aq) --> FeCl2(aq) + H2(g)

When a 7.266-g sample of a particular iron-aluminum alloy was dissolved in excess hydrochloric acid, 0.4849 g of H2(g) was produced. What was the mass percentage of aluminum in the alloy?

Explanation / Answer

from the data

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

2 mole Al = 3 mole H2

Fe(s) + 2 HCl(aq) --> FeCl2(aq) + H2(g)

1 mole Fe = 1 mole H2

no of mole of H2 produced = 0.4849/2 = 0.2424 mole

let us take , total H2 is produced by Al,Fe in 3:1 ratio

so that ,

3/2*(x/27)+(7.266-x)/55.845 = 0.2424

   x = mass of Al in sample = 2.98 g

   mass of Fe = 7.266 - 2.98 = 4.286 g

masspercentage of Al = 2.98/7.266*100 = 41%

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