1 mole of Mg requires 0.5 mole of O2 for oxidation.
Moles of Mg = 30/24
Moles of O2 = 30/32
Moles of O2 required = (1/2)(Moles of Mg) = 30/48
Hence moles of O2 left = 30/32 - 30/48 = 5/16 = 10 g of O2
Moles of MgO formed = 30/24 = (30/24)(40) g = 50 g
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