Question

Determine the mass of iron in 79.2 g of Fe2O3.

Answer 1

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

Answer 2

Answer: The mass of iron present is 55.40 g

Explanation:

We know that:

Molar mass of iron (III) oxide = 159.7 g/mol

Molar mass of iron = 55.85 g/mol

We are given:

Mass of iron (III) oxide = 79.2 g

To calculate the mass of iron in given amount of iron (III) oxide, we use unitary method:

In 159.7 grams of iron (III) oxide, the mass of iron is (2 × 55.85) = 111.7 g

So, in 79.2 grams of iron (III) oxide, the mass of iron will be = $\frac{111.7}{159.7}\times 79.2=55.40g$

Hence, the mass of iron present is 55.40 g