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

1,203.84 grams of FeO will be needed.

Step-by-step explanation:

`4FeO\rightarrow Fe_3O_4 + Fe`

Moles of iron =
`(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:

`(4)/(1)* 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.

Answer 2

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