A granulated sample of aircraft alloy(Al,Mg,Cu) weighing 8.72 g was first treated with alkali to dissolved Al, then with very dilute HCl to dissolve Mg,leaving a residue of Cu. The residue after alkali boiling weighed 2.1 g and the acid insoluble residue from this weighed 0.69g. What is the composition of the alloy?

Asked by Prince Sonu | 18th Apr, 2013, 09:29: PM

Expert Answer:

We know that the amount lost during alkali boiling is the total amount of aluminum: subtract 2.1g from 8.72g to get 6.62g Al. The residue after acid is the mass of the copper: .69g Cu. The mass of Mg is the amount that dissolved in the acid: 2.1g - .69g = 1.41g.

Divide each mass by the total amount of alloy used and multiply by 100% to get the percentage of each element in the alloy.

(6.62/8.72) x 100% = 75.9% Al
(1.14/8.72) x 100% = 13.1% Mg
(0.69/8.72) x 100% = 7.9% Cu

Answered by  | 19th Apr, 2013, 08:59: PM

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