Question

What mass of oxygen reacts when 84.9 g of iron is consumed in the following reaction: Fe+O2= Fe2O3

Answer 1

First we will calculate the number of moles of Iron:

`n = (m)/(M)`, where n is the number of moles, m is the mass of iron in the reaction and M is the Atomic weight.

`n= (84.9)/(55.845) = 1,52` moles of Iron.
The same number of moles of Oxygen will take part in the reaction.
So
`1,52= (mOxygen)/(32)` where 32 is the Atomical Weight of Oxygen (16 x 2).
=>
`mOxygen=32*1,52=48,64`g

Answer 2

36.385 grams of oxygen reacts when 84.9 grams of iron.

Step-by-step explanation:

`4Fe+3O_2\rightarrow 2Fe_2O_3`

Moles of iron =
`(84.9 g)/(56 g/mol)=1.5160 mol`

According to reaction, 4 moles of iron reacts with 3 moles of oxygen gas.

Then 1.5160 moles of iron will react with:

`(3)/(4)* 1.5160 mol=1.1370 mol` of oxygen gas

Mass of 1.1370 moles of oxygen gas:

`1.1370 mol* 32 g/mol=36.385 g`

36.385 grams of oxygen reacts when 84.9 grams of iron.