Question

Fe(s) + O2(g) --> Fe3O4(s) When 13.54 g of O2 is mixed with 12.21 g of Fe, what mass in grams of excess reactant remains when the reaction is complete? 2 decimal places

Answer 1

`8.86"g"` of oxygen

Explanation:

We have the reaction:

`3"Fe"+2"O"_2rarr"Fe"_3"O"_4`

So for every `3` moles of iron that react, `2` moles of oxygen react with it.

Converting both to moles:

Iron: `12.21/55.845=0.219"mol"`

Oxygen: `13.54/31.998=0.423"mol"`

We see that oxygen is in excess. To calculate how many moles of oxygen react with the iron:

`3/2=0.219/x`

`x=(0.219*2)/3`

`x=0.146"mol"`

So, we know that `0.423-0.146=0.277"mol"` of oxygen goes unreacted.

Converting back to mass:

Mass of oxygen: `0.277*31.998=8.86"g"` of oxygen goes unreacted.