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

1 Answer
Mar 3, 2018

#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.