Question

Use the law of constant composition to complete the following table summarizing the amounts of iron and chlorine produced upon the decomposition of the sample of iron(III) chloride. Mass FeCl3 Mass Fe Mass Cl Sample A 3.785 g 1.302 g 2.483 g Sample B 1.475 g _____ _____

Answer 1

Ar(Fe) = 55,85
Ar(Cl) = 35,45
Mr(FeCl3) = 162,2

m(Fe) = m(FeCl3) * (M(Fe)/M(FeCl3)) * n(Fe)
m(Fe) = 1,475 g * (55,85 gmol-1/162,2 gmol-1) * 1 mol
(n (Fe) is 1 mole because FeCl3 is consisted of 1 mole of Fe and 3 moles of Cl)
m(Fe) = 0,5078 g ~ 0,508 g

m(Cl) = m(FeCl3) * M(Cl)/M(FeCl3) * n(Cl)
m(Cl) = 1,475 g * (35,45 gmol-1/162,2 gmol-1) *3
m(Cl) = 0,9671 g ~ 0,967 g