A granulated sample of aircraft alloy (Al, Mg, Cu) weighing 8.72 g was first treated with alkali and then with very dilute HCI, leaving a residue. The residue after alkali boiling weighed 2.10 g and the acid insoluble residue weighed 0.69 g. What is the composition of the alloy?

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Al=75.9%. Mg=16.2%. Cu=7.9%

Al=16.2%. Mg=75.9%. Cu=7.9%

Al=7.9%. Mg=16.2%. Cu=75.9%

Al=75.9%. Mg=19.2%. Cu=4.9%

answer is A.

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Detailed Solution

Let mass of Al, Mg and Cu be a, b and c grams respectively.2Al + 2NaOH + H2O→2NaAlO2 + 3H2Mg + HCl →MgCl2+H2 That is, only Aluminium react with NaOH and only Mg reacts with HCl.Therefore a+b+c=8.72 gramb+c=2.10 gram= residue left after alkali treatment.c=0.69 gram residue left after acid treatmentSo, b+0.69 =2.10 therefore b=1.41gram a+1.41g + 0.69=8.72gramSo, a=8.72−1.41+0.69=6.62gram Percentage of Al = 6.628.72×100 =75.91Percentage of Mg = 1.418.72×100 = 16.17Percentage of Cu = 0.698.72×100 = 7.91

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