Answer to Question #118874 in General Chemistry for brian

Question

Iron bromide (FeBr3), and sulfuric acid (H2SO4) react to form iron (III) sulfate (Fe2(SO4)3) and hydrobromic acid (HBr). You have 33.4 g of iron bromide and 10.2 g of sulfuric acid. Answer the following questions:

What is the balanced equation for this reaction? -1)



If you use all 33.4 g of iron bromide, how many grams of iron (III) sulfate can be formed? Work must be shown to earn credit





If you use all 10.2 g of sulfuric acid, how many grams of iron (III) sulfate can be formed? Work must be shown to earn credit



How many grams of iron (III) sulfate will actually be made? Which reagent is limiting? Explain your answer.

Answer 1

1) $2FeBr_3 + 3H_2SO_4=Fe_2(SO_4)_3+6HBr$

2)

$n(FeBr_3)=33.4/(56+3*80)=0.11284 mol$

$n(Fe_2(SO_4)_3)=0.1184/2=0.0592mol => m(Fe_2(SO_4)_3)=$

$=n(Fe_2(SO_4)_3)*M(Fe_2(SO_4)_3)=0.0592*400=23.68g$

3)

$n(H_2SO_4)=10.2/98=0.1041 mol$

$n(Fe_2(SO_4)_3)=0.1041/3=0.0347mol => m(Fe_2(SO_4)_3)=$

$=n(Fe_2(SO_4)_3)*M(Fe_2(SO_4)_3)=0.0347*400=13.88g$

4)

$n(H_2SO_4)=10.2/98=0.1041 mol$ and sulf. acid requires (stoichiometry):

$n(FeBr_3)=2*0.1041/3=0.0694mol$ of Fe (III) bromide, so sulf. acid is limit reagent (0.1184 mol of iron bromide are available)

It means we'll get 13.88g of iron sulfate (3rd point)