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:
$n = \frac{m}{M}$, where n is the number of moles, m is the mass of iron in the reaction and M is the Atomic weight.
$n= \frac{84.9}{55.845} = 1,52$ moles of Iron.
The same number of moles of Oxygen will take part in the reaction.
So $1,52= \frac{mOxygen}{32}$ where 32 is the Atomical Weight of Oxygen (16 x 2).
=>$mOxygen=32*1,52=48,64$g

Answer 2

Answer:

36.385 grams of oxygen reacts when 84.9 grams of iron.

Explanation:

$4Fe+3O_2\rightarrow 2Fe_2O_3$

Moles of iron = $\frac{84.9 g}{56 g/mol}=1.5160 mol$

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

Then 1.5160 moles of iron will react with:

$\frac{3}{4}\times 1.5160 mol=1.1370 mol$ of oxygen gas

Mass of 1.1370 moles of oxygen gas:

$1.1370 mol\times 32 g/mol=36.385 g$

36.385 grams of oxygen reacts when 84.9 grams of iron.