Answer to Question #144358 in General Chemistry for Rj

Question

Consider the following chemical reaction:

2 AlBr3 + 3 K2SO4 = 6 KBr + 1 Al2(SO4)3

If 306 g AlBr3 is reacted with excess K2SO4 to obtain 118 g KBr, what was the percent yield for this reaction?

Answer 1

2AlBr₃ = 6KBr

2×[27+3(80)]AlBr₃ = 6× (39+80) KBr

2×267g AlBr₃ = 6×119 g KBr

1g AlBr₃ = (6×119)/(2×267) g KBr

306g AlBr₃ = (6×119×306)/(2×267)g KBr

= 409.15g

%yield = actual yield/ theoretical yiedlld ×100%

% yield = 118g/409.15 ×100%

% yield = 28.84%