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
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
Related Videos
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change