Reakcija smese

aristotel27's picture

U reakciji 10,2g smese Mg i Al sa HCl izdvaja se 3*10(23) molekula vodonika.Izracunati procentni(%) sastav smese.

milicaradenkovic's picture

Mg+2HCl-------------------------->MgCl2+H2

2Al+6HCl----------------------------->2AlCl3+3H2

n(H2)=N(H2)/Na

n(H2)=3*10(23)/6*10(23)=0.5 mol

n(H2)ukupno=n(H2)1+n(H2)2

m(Al)+m(Mg)=10.2 g

n(Al):n(H2)2=2:3

n(H2)2=3*n(Al)/2

n(Mg):n(H2)1=1:1

n(Mg)=n(H2)1

n(Al)*Mr+n(Mg)*Mr=10.2

n(Mg)+3*n(Al)/2=0.5

27*n(Al)=10.2-24*n(Mg)

n(Al)=(10.2-24*n(Mg))/27

n(Mg)+3*(10.2-24*n(Mg))/27*2=0.5

54*n(Mg)+30.6-72*n(Mg)=27

18*n(Mg)=3.6

n(Mg)=3.6/18=0.2 mol

n(H2)2=n(H2)ukupno-n(H2)1

n(H2)2=0.5-0.2=0.3 mol

n(Al)=2*0.3/3=0.2 mol

m(Al)=n(Al)*Mr=27*0.2=5.4 g

m(Mg)=n(Mg)*Mr=24*0.2=4.8 g

w(Mg)=m(Mg)/m(uzorka)=4.8/10.2=0.4706=47.06 %

w(Al)=m(Al)/m(uzorka)=5.4/10.2=0.5294=52.94 %

 

 

 

 

 

aristotel27's picture

hvala

aristotel27's picture

samo mi nije jasno ovo:n(Mg)+3*n(Al)/2=0.5?

milicaradenkovic's picture

Tako sam izrazila molove vodonika iz obe reakcije jer je 0.5 ukupna kolicina vodonika.

aristotel27's picture

ok,a ova racunica nemogu da je smislim xD

54*n(Mg)+30.6-72*n(Mg)=27

18*n(Mg)=3.6

milicaradenkovic's picture

72*n(Mg)-54*n(Mg)=30.6-27 i dobijes

18*n(Mg)=3.6