A 55.0g sample of iron (III) filings is reacted with 23.8g of powdered sulfur (S8). How much iron (III) sulfide in moles would be produced in this reaction?

Equation:

Convert to moles
of iron:

Convert to moles
of sulfur:

Calculate the
limiting reagent:

Solve the problem:

Question

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Respuesta :

blahblahmali blahblahmali · Answer 1 · 2019-05-16T06:39:06+02:00

Answer:

0.744 mol

Explanation:

the balanced equation for the reaction is

8Fe + S₈ ---> 8FeS

molar ratio of Fe to S₈ is 8:1

number of moles of Fe - 55.0 g / 56 g/mol = 0.98 mol

number of moles of S - 23.8 g / 256 g/mol = 0.093 mol

if we are to assume that S₈ is the limiting reactant

if 1 mol of S₈ reacts with 8 mol of Fe

then 0.093 mol of S₈ reacts with - 8 x 0.093 mol = 0.744 mol of Fe

however there's 0.98 mol of Fe present but only 0.744 mol of Fe is needed

therefore Fe is in excess and S₈ is the limiting reagent

molar ratio of S₈ to FeS is 1:8

then 0.093 mol of S₈ reacts with - 8 x 0.093 = 0.744 mol of FeS

number of FeS moles produced is 0.744 mol

Oseni Oseni · Answer 2 · 2020-02-27T12:10:11+01:00

Answer:

0.74 moles iron (III) sulfide

Explanation:

From the balanced equation of reaction:

[tex]Fe + S --> FeS[/tex]

1 mole of Fe reacts with 1 mole of S to give 1 mole of FeS.

moles = [tex]\frac{mass}{molar mass}[/tex]

mole of Fe = 55/55.8 = 0.99 moles

mole of S = 23.8/32.07 = 0.74 moles

Sulfur is limited in quantity and will therefore determine the rate of reaction.

1 mole of sulfur gives 1 mole of FeS

0.74 moles of sulfur will therefore give 0.74 moles of FeS.

0.74 moles iron (III) sulfide will be produced.