1 mole of KClO3 is thermally decomposed and excess of aluminium is burnt in the gaseous product . how many moles of Al2O3 are formed?? sir i did not understand the meaning of the question. please explain what the question is looking for and the steps of the answer. thank you.
Asked by Sayoni Maiti
| 29th May, 2014,
09:20: AM
The thermal decomposition of KClO? and the subsequent oxidation of Aluminum to form Al?O?.
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. Hence,
1.5 mol O? x ( 2 mol Al?O? / 3 mol O?) = 1 mol Al?O?
The thermal decomposition of KClO? and the subsequent oxidation of Aluminum to form Al?O?.
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. Hence,
1.5 mol O? x ( 2 mol Al?O? / 3 mol O?) = 1 mol Al?O?
Answered by Prachi Sawant
| 29th May, 2014,
11:07: AM
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