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If 279 grams of iron react with an excess of oxygen, as shown in the balanced equation below, how many grams of iron (III) oxide can be formed? 4Fe + 3O2 2Fe2O3
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@Abmon98
Number of Moles=Mass(g)/Molar Mass(g/mol) 279/56=4.98 moles of iron Look for the ratio of moles of iron to iron oxide which is 4:2 4.98*2/4=2.49 moles 2.49=Mass/159 Mass=2.49*156=369.0grams
You mean399?
sorry the answer is 395.91
139.5 grams 199.5 grams 399 grams 1596 grams No?.. It has to be 399
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(279/56)*2/4*159.69=397.799 grams
.-. So 399 still.... Because that's the only one that fits
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