Question

2 Al(s) + 6 HCl(aq) --> 2 AlCl3(aq) + 3 H2(g) Fe(s) + 2 HCl(aq) --> FeCl2(aq) + H2(g) When a 7.007-g sample of a particular iron-aluminum alloy was dissolved in excess hydrochloric acid, 0.4705 g of H2(g) was produced. What was the mass percentage of aluminum in the alloy?

Answer 1

Fe +2 HCl --> FeCl2 + H2
1 mole Fe forms 1 mole H2

2 Al + 6 HCl --> 2 AlCl3 + 3 H2  
or, Al+ 3HCl------> AlCl3+3/2 H2
1 mole Al forms 3/2 moles H2

Given that;

0.4705 g of H2(g) was produced

Moles of H2 = 0.4705 g of H2 / 2.01588 g/ moles

=0.233 moles H2

Let there are x moles of Fe and y moles of Al in the alloy

Hydrogen balance

x+1.5y = 0.233 ----------------[ 1 ]

x = 0.233 -1.5 y

Molar mass of Fe =55 .85 g / mol

Molar mass of Al = 27 g / mol
Also we have,
55.85 x+27y = 7.007 ------------[ 2 ]

now Solve eqn [1] and [2],

55.85 x+27y = 7.007

55.85 (0.233 -1.5 y)+27y = 7.007

13.01 – 83.775 y +27 y = 7.007

13.01 – 56.775 y = 7.007

56.775 y =13.01 – 7.007

56.775 y = 6.003

y = 0.106

and

x = 0.233 -1.5 y

x = 0.233 -1.5 *0.106

= 0.39

Amount of Al = 0.106 moles*27 g/ mole

= 2.862 g


% Al in alloy = amount of Al / amount of alloy *100

= 2.862/ 7.007*100

= 40.84 %