Calculating the mass of iron (II) oxide produced from 56.8 g of iron in excess to oxygen

Question

help please

I balanced the equation

Fe+O2 = FeO

so 2Fe+1O2= 2FeO

now what?

Answer 1

It's great that you are starting with the balanced chemical reaction equation!! You need this to know the mole-to-mole ratio (converting the "Fe" to "FeO").

"in excess oxygen" means you don't have to worry about running out of oxygen, you have plenty to convert all the "Fe" into "FeO".

To do this problem, first calculate the molar mass of the compounds involved. You need "Fe" (what you are starting with), which you can just look up in the Periodic Table. It is 55.85 g/mol. Then you need "FeO" (what you are producing) which consists of 1 atom of "Fe" at 55.85 plus 1 atom of "O" at 15.999 which adds to 71.84 g/mol.

Now you can use a method like this to lay out the conversions:

You start on the left with the grams of "Fe", convert to moles using the molar mass, then use the molar ratio from the balanced chemical reaction equation to convert to moles of "Fe" to moles of "FeO", then use the molar mass to convert to grams of "FeO". Notice that I've canceled out the units that are the same on top and bottom, leaving you with grams of FeO (which is what you are seeking).

Multiply everything on top, divide everything on bottom:

(56.8•1•2•71.84)/(55,85•2•1) = 73.1 g of FeO