Koliko grama supstituisanog proizvoda moze nastati u reakciji 0,05 mola fenola sa dovoljnom kolicinom broma?RESENJE JE 16,55G.
C6H5OH+3Br2------------->C6H2OHBr3+3HBr
n(C6H2OHBr):n(fenola)=1:1
n(C6H2OHBr3)=n(fenola)=0.05 mol
m(C6H2OHBr3)=n(C6H2OHBr3)*Mr=0.05*331=16.55 g
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fenoli
C6H5OH+3Br2------------->C6H2OHBr3+3HBr
n(C6H2OHBr):n(fenola)=1:1
n(C6H2OHBr3)=n(fenola)=0.05 mol
m(C6H2OHBr3)=n(C6H2OHBr3)*Mr=0.05*331=16.55 g