What is the maximum number of grams of Iron (III) sulfide, Fe2S3, which can be obtained...

What is the maximum number of grams of Iron (III) sulfide, Fe₂S₃, which can be obtained from the reaction of 80.0g of Fe and 78.0g of S?

(FW Fe=55.85, S=32.07, Fe₂S₃=207.91) Please show work

Solutions

Expert Solution

2Fe + 3S ------> Fe2S3

2 moles 3 moles

no of moles of Fe = W/A.Wt

= 80/55.85 = 1.43 moles of Fe

no of moles of S = 78/32.07 = 2.43 moles of S

from balanced equation

2 moles of Fe react with 3 moles of S

1.43 moles of Fe react with = 3*1.43/2 = 2.145 moles of S requid So it is excess reagent .Limiting reagent is Fe

2 moles of Fe react with S to form 1 moles of Fe2S3

1.43 moles of Fe react with S to form = 1*1.43/2 = 0.715 moles of Fe2S3

mass of Fe2S3 = no of moles of Fe2S3* Gram molar mass

= 0.715*207.91 = 148.65565 gm of Fe2S3

queen_honey_blossom answered 1 year ago

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