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Question

3.90 g of a mixture of Al and Al2O3, when reacted with a solution of NaOH produced 840 mL of gas at NTP. Find % of Al2O3 in mixture (as nearest integer).

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Solution

The reactions that are involved are:
2Al(s)+2NaOH+6H2O2Na++2[Al(OH)4]+3H2(g)
2 moles 3 moles gas
Al2O3+2NaOH+3H2O2Na++2[Al(OH)4]
So, mass of Al in sample =2×27×8403×22400=0.675 g
So, % of Al =0.675×1003.9=17.3 and
% of Al2O3 in mixture =83

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