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# One mole of kclo3 is heated in presence of mno2 . The produced oxygen is used in burning of Al . Then oxide of Al that will be formed

Umakant biswal
5359 Points
5 years ago
@ sree varsha
Two reactions occur here: the thermal decomposition of KClO? and the subsequent oxidation of Aluminum to form Al?O?. Our two reactions are:

2 KClO? ? 2 KCl + 3 O? and 4 Al + 3 O? ? 2 Al?O
The molar ratio for the first reaction is 2:3 for KClO?: O?. Thus, if 1 mole KClO? is decomposed, 1.5 moles of O? is produced ===> 1 mol KClO? x ( 3 mol O? / 2 mol KClO?) = 1.5 mol O?
The molar ratio for the second reaction is: 3:2 for O? : Al?O?. ===> 3 moles of oxygen produces 2 moles of aluminum oxide. Since we have 1.5 moles of oxygen (half of the three needed) we can expect to produce half of the 2 moles or 1 mole aluminum oxide....Here's the math
1.5 mol O? x ( 2 mol Al?O? / 3 mol O?) = 1 mol Al?O?