- #1

JonnyG

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## Homework Statement

Suppose Iron(II) Sulfide is reacted with hydrogen chloride. How many grams of HCl is required is react with 75.0 grams of Iron (II) Sulfide ore of which 30% is inactive.

## Homework Equations

## The Attempt at a Solution

EDIT: Nevermind I solved it. Obviously 43.6 g of [itex] 2 \mathrm{HCl} [/itex] is the same as 43.6 g of [itex] \mathrm{HCl} [/itex].

Only 70% of the Iron (II) Sulfide is active, so we can just pretend we are dealing with 52.5 g of Iron (II) Sulfide. Now, the equation is [itex] \mathrm{FeS} + 2\mathrm{HCl} \rightarrow \mathrm{FeCl}_2 + \mathrm{H}_2 \mathrm{S} [/itex].

1 mole of [itex] \mathrm{FeS} = 87.95 [/itex] g and 1 mole of [itex] 2\mathrm{HCl} = 73 [/itex] g. Because 52.5/87.95 = 0.597 then we only need (0.597)(73) = 43.6 g of [itex] 2 \mathrm{HCl} [/itex] for the reaction, which is equivalent to (43.6)(2) = 87.2 g of [itex] \mathrm{HCl} [/itex]. Is this correct? A couple answers that I read online says that the answer is 43.6 g of [itex] \mathrm{HCl} [/itex] rather than 43.6 g of [itex] 2 \mathrm{HCl} [/itex].