Asked by megan
How many moles of iron(II)sulfide are produced when 4.38g of Fe reacts with 2.88g of S to produce iron(II)sulfide?
Answers
Answered by
DrBob222
This is a limiting reagent problem.
Write the equation and balance it.
16Fe + 3S8 ==> 8Fe2S3
Convert 4.38 g Fe to mols. mols = g/molar mass.
Convert 2.88 g S8 to mols S8.
a.Using the coefficients in the balanced equation, convert mols Fe to mols Fe2S3.
b. Same thing but convert mols S8 to mols Fe2S3.
c. The smaller number of mols of Fe2S3 is the correct one to choose and the reactant that produced that is the limiting reagent.
The number in c x molar mass = grams Fe2S3.
Post your work if you get stuck.
Write the equation and balance it.
16Fe + 3S8 ==> 8Fe2S3
Convert 4.38 g Fe to mols. mols = g/molar mass.
Convert 2.88 g S8 to mols S8.
a.Using the coefficients in the balanced equation, convert mols Fe to mols Fe2S3.
b. Same thing but convert mols S8 to mols Fe2S3.
c. The smaller number of mols of Fe2S3 is the correct one to choose and the reactant that produced that is the limiting reagent.
The number in c x molar mass = grams Fe2S3.
Post your work if you get stuck.
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