Question

Fe (s) + S (I) --> Fes (s) In the experiment above, 7.62 g of Fe
are allowed to react with 8.67 g of S. How much FeS is
formed?
a) 14.8 g
b) 12.2 g
C) 13.7 g
d) 19.9 g

Answer 1

The amount of FeS which is formed in the given reaction is 12.2 grams.

How do we convert mass into moles?

Relation between the mass and moles of any substance will be shown as:
n = W/M, where

• W = given mass
• M = molar mass

Moles of Fe = 7.62g / 55.8g/mol = 0.136mol

Moles of S = 8.67g / 32g/mol = 0.2709 = 0.271mol

From the stoichiometry of the given chemical reaction it is clear that:

• 1 mole of Fe reacts with 1 mole of S.

• It means 0.271 moles of sulfur required 0.271 moles of iron.
• But iron is not present in sufficient amount so it is the limiting  reagent.

Formation of product depends on the amount of Fe, so 0.136 moles of Fe produces 0.136 moles of FeS.

Now mass of FeS = (0.136mol)(87.91g/mol) = 11.95g = 12g (12.2 approx)

Hence required mass of FeS is 12.2 grams.

To know more about mass & moles, visit the below link:
https://brainly.com/question/15373263

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