Koliko se molekula broma adira na 1 molekul etina, ako je poznato da bromovanjem 5,2g etina nastaje 69,2g bromovanog derivata?
m(Br2)=69.2-5.2=64g n(Br2)=64g/160g/mol=0.4mol N(Br2)=0.4*6*10^23 n(C2H2)=5.2g/26g/mol=0.2mol N(C2H2)=0.2*6*10^23 (0.4*6*10^23):(0.2*6*10^23)=x:1 x=2
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m(Br2)=69.2-5.2=64g n(Br2)=64
m(Br2)=69.2-5.2=64g
n(Br2)=64g/160g/mol=0.4mol
N(Br2)=0.4*6*10^23
n(C2H2)=5.2g/26g/mol=0.2mol
N(C2H2)=0.2*6*10^23
(0.4*6*10^23):(0.2*6*10^23)=x:1
x=2