Calculating Iron (II) Sulfide: Stoichiometry Help & Equation Explanation

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SUMMARY

The discussion focuses on calculating the amount of Iron (Fe) required to produce 56.8 grams of Iron (II) Sulfide (FeS) using stoichiometry. The correct molar ratio between Iron and Iron (II) Sulfide is established as 1:1. The molar masses are noted as Fe = 55.8 g/mol and S = 32.1 g/mol, leading to a total molar mass of FeS as 87.9 g/mol. The calculation confirms that 36.1 grams of Iron is needed to produce the desired amount of Iron (II) Sulfide.

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nduncan2010
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stoichiometry--hellp please!

1. How much Iron should be combined with sulfur to produce 56.8 grams of iron (II) Sulfide



2.



3. ok.. so this is what i have ..
Fe + S ----> FeS ----correct??

Then. molar ratios are 1-1-1..
The molar ration between Fe and FeS is 1:1
Fe-55.8
S-32.1
--------
= 87.9

56.8 /87.9 *1*55.8 = 36.1 g FeS.. is this correct? the wording in the equation is different then all of the other equations on my worksheet soo I am not sure if this is right.
 
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Your type-setting or your text formatting is not good. You want to show as:
(56.8 /87.9) *1*55.8 = 36.1 g Fe
That then would be good.

NOTE: why the forum does not allow text strike-out? I meant to show previous post's answer for FeS which was wrong, and then make a 'scratch' mark through the S; because the 36.1 grams is for the IRON.
 
Last edited:


ohhh yeahh . I messed the label up... oops.. thanks for catching that!
 

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