1.12 g of iron reacts with oxygen to form 1.60 g of an oxide of iron. use relative atomic masses: fe = 56, o = 16. what is the formula of this oxide of iron?
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n of Fe = 1.12/56 = 0.02 mol
Mass of oxygen in Oxide = 1.60g-1.12g = 0.48g
So n of Oxygen = 0.48/16 = 0.03 mol
molar ration of Fe:O is therefore 0.02:0.03
Dividing each by the smallest value (0.02) gives a ratio of 1 : 1.5
As we need an integer we must multiply both values by 2, giving an empirical ratio of Fe to Oxygen as 2:3 Thus the answer is D - Fe2O3
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n of Fe = 1.12/56 = 0.02 mol
Mass of oxygen in Oxide = 1.60g-1.12g = 0.48g
So n of Oxygen = 0.48/16 = 0.03 mol
molar ration of Fe:O is therefore 0.02:0.03
Dividing each by the smallest value (0.02) gives a ratio of 1 : 1.5
As we need an integer we must multiply both values by 2, giving an empirical ratio of Fe to Oxygen as 2:3 Thus the answer is D - Fe2O3
plzzz mark as brainliest
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