3 and 4 are subscripts. Please explain the method.
Asked by Sayoni Maiti
| 18th Apr, 2014,
07:20: PM
Given:
Two metallic oxides contain 27.6% and 30% oxygen respectively.
Ratio of metal and oxygen in first oxide, M3O4 = 72.4 : 27.6
Ratio of metal and oxygen in second oxide = 70 : 30
Let molecular mass of metal = M
Therefore, the percentage by weight of the metal in the oxide

Now,
Moles of metal in second oxide = 70 / 56 = 1.25
Moles of oxygen in second oxide = 30 / 16 = 1.875
Ration of moles of metal and oxygen = 1.25 : 1.875
= 1 : 1.5
= 2 : 3
Hence, formula of the second oxide = M2O3
Given:
Two metallic oxides contain 27.6% and 30% oxygen respectively.
Ratio of metal and oxygen in first oxide, M3O4 = 72.4 : 27.6
Ratio of metal and oxygen in second oxide = 70 : 30
Let molecular mass of metal = M
Therefore, the percentage by weight of the metal in the oxide
Now,
Moles of metal in second oxide = 70 / 56 = 1.25
Moles of oxygen in second oxide = 30 / 16 = 1.875
Ration of moles of metal and oxygen = 1.25 : 1.875
= 1 : 1.5
= 2 : 3
Hence, formula of the second oxide = M2O3
Answered by
| 21st Apr, 2014,
12:33: PM
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