Question

The low-grade iron ore taconite, which contains Fe3O4, is concentrated and made into pellets for processing. If one ton of taconite pellets yields 545 lb. of iron, what percentage of the pellets is Fe304?

Answer 1

Answer: The percent composition of
`Fe_3O_4` in taconite is 37.6 %.

Step-by-step explanation:

We are given:

Mass of taconite pellets = 1 ton = 907185 g (Conversion factor: 1 ton = 907185 g)

Mass of iron produced = 545 lb = 247212 g (Conversion factor: 1 lb = 453.6 g )

We know that:

Molar mass of iron = 55.85 g/mol

Molar mass of
`Fe_3O_4` = 231.53 g/mol

1 mole of
`Fe_3O_4` contains 3 moles of iron atom and 4 moles of oxygen atom

(3 × 55.85) = 167.55 g of iron is produced from 231.53 grams of
`Fe_3O_4`

So, 247212 grams of iron will be produced from =
`(231.53)/(167.55)* 247212=341611.43g` of
`Fe_3O_4`

To calculate the percentage of
`Fe_3O_4` in taconite, we use the equation:

`\%\text{ composition of }Fe_3O_4=\frac{\text{Mass of }Fe_3O_4}{\text{Mass of taconite}}* 100`

Mass of taconite = 907185 g

Mass of
`Fe_3O_4` = 341611.43 g

Putting values in above equation, we get:

`\%\text{ composition of }Fe_3O_4=(341611.43g)/(907185g)* 100=37.6\%`

Hence, the percent composition of
`Fe_3O_4` in taconite is 37.6 %.