# Chemistry

When iron (Fe) rusts (reacts with O2), the product is iron(III) oxide (Fe2O3).
If 25 g Fe is reacted with 25 g O2 in a sealed container, what are the masses of Fe, O2 and Fe2O3 in the container after the reaction is complete?

How is type of question calculated?

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1. substances react by molar quantities ... so you need to convert to moles

find the moles of Fe and O2 ... by dividing by the molar masses

two Fe and three O react to form the product

there will be excess O2 remaining ... all the Fe will be reacted

find the moles of Fe2O3 formed , and the moles of O2 remaining

convert the moles to masses ... by multiplying by the molar masses

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2. First, determine the reaction equation. You probably want
4Fe + 3O2 = 2Fe2O3
Fe is 55.8 g/mole
O is 16 g/mole
so,
25g Fe = 0.448 moles
25g O2 = 0.781 moles
In the reaction, Fe:O2 = 4:3
so, 0.448 moles Fe (0.224 moles of Fe2) requires 0.336 moles of O2
So there is not enough Fe to consume all the O2. There will be 0.445 moles of O2 left over.
So, the masses after the reaction are
Fe: 0
O2: 0.445*32 = 14.25g
Fe2O3: 0.224 * 159.6 = 35.75 g
Note that the total mass is still 50g

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oobleck
3. Here can you please tell me that can i use thus equation for rusting of iron Fe2O3 +3CO ----- 2Fe +3CO2

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