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

Atomic weights : Al = 27 and Fe= 56

Let x= mass of Al in the alloy.

Moles of Al in the alloy = x/27, mass of Fe= 7.266-x, moles of Fe= (7.266-x)/56

From the reaction . 2 Al(s) + 6 HCl(aq) --> 2 AlCl3(aq) + 3 H2(g)

2 moles of Al produced 3 moles of H2, i.e 6 g of H2( molar mass of H2=2)

x/27 moles produces 6x/2*27 gm of H2=x/9 gm of H2

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

1 mole of Fe produces 1 mole of H2. i.e 2 gm of H2

(7.266-x)/56 moles of Fe produces (7.266-x)*2/56= (7.266-x)/28 gm of H2

Total grams of H2 produced = x/9+(7.266-x)/28 = 0.4849

0.111x+0.2595-0.036x= 0.4849

Hence x*(0.111-0.036)= 0.4849-0.2595

Hence x= 3 gm , mass % of Al in the alloy =100*3/7.266 =41.29%

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