Determine the mass of iron in 79.2 g of Fe2O3.

55.4-g Fe.

Begin by balancing the equation:

2 Fe2O3 - > 4 Fe + 3 O2

Start your stoic:

(79.2g Fe2O3) < - - you start with what you have and always try to get rid of what unit you are at

Conversion to get rid of grams is

# grams over mole

^

This number comes from periodic table

Add whole mass of Fe2O3 = 159.697g

(79.2g Fe2O3) (. mole.)

(159.697g)

Now get rid of moles, this is where you used the balanced numbers at the beginning

The number you are looking for goes on top-> (4 moles Fe)

(2 moles Fe2O3)

Last get mass of Fe alone and plug in to calc

Whole equation:

(79.2g) (moles) (4mFe) (55.85gFe)

(159.697) (2mole) (mole)

79.2 divides by 159.697 multiplied by 4 divided by 2 multipled by 55.85 equals 55.396406 rounded to:

55.4 g Fe