Respuesta :
1.0 mol of mg is mixed with 2.0 mol of br2, and 0.84 mol of mgbr2 is obtained the percentage yield of the reaction is 84%
mg(s) + Br[tex]_{2}[/tex](l) ⇒ mgBr[tex]_{2}[/tex](l)
1 mol 2mol 0.84
According to balance chemical equation given above one molecule of mg react with one molecule of Br[tex]_{2}[/tex]. Hence 1 mol of mg will react with 1 mol of
Br[tex]_{2}[/tex]. mg will finish first in the react. so mg will be limiting reagent for this reaction.
% yield = [tex]\frac{obsered yield}{Theoritical yield}[/tex] × 100
Observed yield = 0.84 mol = o.84 mol×184.11 g/mol of mgBr[tex]_{2}[/tex]
= 154.65 g
Theoritical yield( Amount of product expected in ideal condition ) = 1 mol
= 1 mol×184.11 g/mol = 184.11 g
%yield = [tex]\frac{154.65 g}{184.11 g}[/tex]× 100 = 84%
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