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

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To produce 56.8 grams of iron (II) sulfide (FeS), the correct stoichiometric calculation involves combining iron (Fe) and sulfur (S) in a 1:1 ratio. The molar mass of Fe is 55.8 g/mol and that of S is 32.1 g/mol, leading to a total molar mass of FeS at 87.9 g/mol. The calculation for the mass of iron needed is (56.8 g FeS / 87.9 g/mol) * 55.8 g/mol, resulting in approximately 36.1 grams of iron. The initial confusion arose from a labeling error in the equation, which was clarified in the discussion. Accurate stoichiometric calculations are essential for proper chemical reactions.
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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.
 
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ohhh yeahh . I messed the label up... oops.. thanks for catching that!
 
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