Question

After reacting all of the `FeS` in a `1.5\ L` of a reaction solution that had `[FeS]=0.2\ M`, how many grams of `Fe_2O_3` were formed? `" "` `` `4FeS+7O_2\rightarrow 2Fe_2O_3+4SO_2`

Answer 1

`"24 g"`

Explanation:

It’s given that

• `["FeS"] = "0.2 M"`
• `"Volume of solution = 1.5 L"`

`"Molarity" = "Moles of solute"/"Volume of solution (in litres)"`

`"0.2 M" = "n"/"1.5 L"`

`"n = 0.2 M × 1.5 L = 0.3 mol"`

So, we have `"0.3 mol"` of `"FeS"` in given solution.

`underbrace("4FeS") + "7O"_2 -> underbrace("2Fe"_2"O"_3) + "4SO"_2`
`color(blue)("4 mol") color(white)(................)color(blue)("2 mol")`

From above equation, `"4 mol"` of `"FeS"` reacts to give `"2 mol"` of `"Fe"_2"O"_3`

We have `"0.3 mol"` of `"FeS"`. So, moles of `"Fe"_2"O"_3` formed will be

`0.3 cancel"mol FeS" × ("2 mol Fe"_2"O"_3)/(4 cancel"mol FeS") = "0.15 mol Fe"_2"O"_3`

Molar mass of `"Fe"_2"O"_3` is `"160 g/mol"`

Mass of `"Fe"_2"O"_3` formed `= 0.15 cancel"mol" × "160 g"/cancel"mol" = color(red)"24 g"`