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)
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