Tính hàm lượng % theo khối lượng của các kim loại trong các oxit sau: a) FeO ; ​b) CaO ; ​c) MgO ; ​d) Cu2O

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13 protons and 13 electrons

Unbalanced equation: fe2o3+ cofe + co21 fe2o3 + 3 co = 2 fe(s) + 3 co212.8g fe2o3 x (1 mol fe2o3 / 159.7g fe2o3) = 0.08015 mol fe2o311.5g co x (1 mol co / 28.01g co) = 0.4106 mol cofrom the balanced equation, 1 fe2o3 reacts with 3 cosince we have 0.08015 mol fe2o3 we therefore need0.08015 mol fe2o3 x (3 mol co / 1 mol fe2o3) = 0.2404 mol cosince we have 0.4106 mol co but only need 0.2404 to completely consume all thefe2o3, we have xs (excess) co and fe2o3 is therefore the limiting reagentfrom the balanced equation, 1 fe2o3 = 2 fe(s)0.08015 mol fe x (2 mol fe(s) / 1 mol fe2o3) = 0.1603 mol fe(s)theoretical mass of fe .1603 mol fe x (55.84g fe / mole fe) = 8.951g fe% yield = (6.10g / 8.951g) x 100% = 68.1%i didn't do this work, so i am not one hundred percent sure if it is correct  seems like a reliable website, 

Bbecause it is a psychical reaction.