How Many Grams of HCl Are Needed to React with Iron(II) Sulfide?

[JonnyG]

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 2 \mathrm{HCl} is the same as 43.6 g of \mathrm{HCl}.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 \mathrm{FeS} + 2\mathrm{HCl} \rightarrow \mathrm{FeCl}_2 + \mathrm{H}_2 \mathrm{S}.

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

 

[BvU]

Well, on a scale it comes down to exactly the same quantity of $HCl$