Question

How many grams of FeO would be needed to make 234.2 grams of Fe in the chemical reaction 4FeO = Fe3O4 + Fe? Show your work.

Answer 1

The grams  of FeO  that would  be needed to make 234.2 grams  of Fe  is

1204.42 grams

calculation

4 FeO → Fe₃O₄ +Fe

Step 1: find the  moles  of  fe

moles  =  mass /molar mass

from periodic table the  molar mass of Fe  = 56 g/mol

moles = 234.2 g/56 g/mol = 4.182  moles

Step 2: use  the mole ratio to determine the moles of  FeO

FeO: Fe  is 4:1 therefore the  moles of FeO  =4.182 moles  x4 =16.728  moles

Step 3:   find  the mass of FeO

mass = moles  x molar  mass

The molar mass of FeO = 56 +16 = 72 g /mol

mass  = 16.728 moles  x 72 g/mol= 1204.42 grams

Answer 2

Answer:

1,203.84 grams of FeO will be needed.

Explanation:

$4FeO\rightarrow Fe_3O_4 + Fe$

Moles of iron = $\frac{234.2 g}{56 g/mol}=4.18 mol$

According to reaction, 1 mol of iron is obtained from 4 moles of ferrous oxide.

Then 4.18 moles of iron will be obtained from:

$\frac{4}{1}\times 4.18 mol=16.72 mol$ of ferrous oxide

Mass of 16.72  moles of ferrous oxide:

16.72 mol × 72 g/mol = 1,203.84 g

1,203.84 grams of FeO will be needed.