A 10.0 g sample of a pure metal 'M' (atomic mass 56 g/mol ) is completely oxidized by heating in oxygen to form a mixture of its two oxides, MO and M 2 O 3 . The total mass of the oxide

Question

A 10.0 g sample of a pure metal 'M' (atomic mass 56 g/mol ) is completely oxidized by heating in oxygen to form a mixture of its two oxides, MO and M 2 ​ O 3 ​ . The total mass of the oxide mixture obtained is 13.2 g . What is the mole ratio of MO to M 2 ​ O 3 ​ in the final mixture?

TestHub · Accepted Answer

The mass of oxygen incorporated is 13.2 g − 10.0 g = 3.2 g . Moles of oxygen atoms = 3.2 g /16 g/mol = 0.2 mol . Moles of metal M = 10.0 g /56 g/mol = 5/28 mol . Let x be moles of MO and y be moles of M₂O₃. Metal balance: x + 2 y = 5/28 . Oxygen balance: x + 3 y = 0.2 . Solving these equations gives y = 3/140 mol and x = 19/140 mol . The mole ratio x : y is 19 : 3 .

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