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1.12 g of iron reacts with oxygen to form an oxide.the mass of oxide formed is 1.44g.Formula of oxide is
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1.12 g of iron reacts with oxygen to form an oxide.the mass of oxide f...
As there is no amount of O2 given hence it is not limitingmoles of Fe = 1.12/56= 0.02let an arbitrary rxn4Fe + yO2--> 2Fe2O(y) moles of product formed = half the moles of Fe usedmoles of product = 0.01molar mass of product= mass formed / moles formedmolar mass = 1.44 / 0.01 = 144molar mass = 2* molar mass of Fe + y * molar mass of Oxygen ==> 2*56 + (y)*16 = 144==> 112 + 16 y= 144==> 16y = 32==> y = 2==> now formula of product= Fe2O2== FeO and the rxn become 2Fe + O2 --> 2 FeO
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1.12 g of iron reacts with oxygen to form an oxide.the mass of oxide formed is 1.44g.Formula of oxide is
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