Koliko ce se mol-atoma srebra izdvojiti u reakciji srebrenog ogledala,ako je u reakcij stupilo 44 g 10% etanala?
CH3CHO+2{Ag(NH3)2}+ +2OH- ---------->CH3COOH+2Ag+4NH3+H2O
m(etanala)=mr*w(metanala)=44*0.1=4.4 g
n(etanala)=m(etanala)/Mr=4.4/44=0.1 mol
n(Ag):n(etanala)=2:1
n(Ag)=n(etanala)*2/1=0.1*2/1=0.2 mol
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CH3CHO+2{Ag(NH3)2}+ +2OH- ---------->CH3COOH+2Ag+4NH3+H2O
m(etanala)=mr*w(metanala)=44*0.1=4.4 g
n(etanala)=m(etanala)/Mr=4.4/44=0.1 mol
n(Ag):n(etanala)=2:1
n(Ag)=n(etanala)*2/1=0.1*2/1=0.2 mol