Question

Extraction of Fe (s) from pyrite, FeS2, is a two-step process, as represented above. The maximum amount of Fe (s) that could be extracted from 120 grams of FeSz (molar mass 120 g/mol) is approximately

Answer 1

The maximum amount of Fe that could be extracted from 120 grams of
`FeS_2` is approximately 56 grams. Option B is the right choice.

The given information describes a two-step process for extracting Fe (s) from
`FeS_2` (pyrite). However, the specific steps are missing from the prompt.

To determine the maximum amount of Fe obtainable from 120 grams of
`FeS_2`, we need the complete equation for the first step.

Assuming the first step involves roasting
`FeS_2` with oxygen, the balanced equation would be:

4
`FeS_2`(s) + 11
`O_2`(g) ->
`2 Fe_2O_3`(s) + 8
`SO_2`(g)

From this equation, we can calculate the theoretical yield of
`Fe_2O_3`obtained from 120 grams of
`FeS_2`:

Moles of
`FeS_2` = 120 g / 120 g/mol = 1 mol

Moles of
`Fe_2O_3` produced = 1 mol
`FeS_2` * (2 mol Fe2O3 / 4 mol
`FeS_2`) = 0.5 mol

Theoretical yield of
`Fe_2O_3` = 0.5 mol * 160 g/mol = 80 g

The second step involves reducing
`Fe_2O_3` to Fe using carbon monoxide (CO) as described by the equation:

`Fe_2O_3`(s) + 3 CO(g) -> 2 Fe(s) + 3
`CO_2`(g)

Here, we can assume that all the
`Fe_2O_3` produced in step 1 is converted to Fe in step 2. To calculate the theoretical yield of Fe:

Moles of Fe produced = 0.5 mol
`Fe_2O_3` * (2 mol Fe / 1 mol
`Fe_2O_3`) = 1 mol

Theoretical yield of Fe = 1 mol * 56 g/mol = 56 g

Option B is the right choice.