Procentni sastav

Provincijalka's picture

Dato je 0,640kg smjese C8H18 i C9H20,koji reaguju sa O2 do H20  Ode ceaga se izgradi 0,904kg H2O.Izracunati procentni sastav smjese (Ar: c=12,011; H=1,008; O=15,9994).

milicaradenkovic's picture

2C8H18+25O2------------------->16CO2+18H2O

C9H20+14O2--------------------->9CO2+10H2O

mu(H2O)=m1(H2O)+m2(H2O)

0.904=m1(H2O)+m2(H2O)

m(smese)=m(C8H18)+m(C9H20)

n1(H2O)=18*n(C8H18)/2

n2(H2O)=10*n(C9H20)

m1(H2O)/Mr(H2O)=18*m(C8H18)/2*Mr(C8H18)

m1(H2O)=18*18.0154*10(-3)*m(C8H18)/2*114.232(-3)

m1(H2O)=1.42*m(C8H18)

m2(H2O)=1.405*m(C9H20)

0.904=1.42*m(C8H18)+1.405m(C9H20)

0.640=m(C8H18)+m(C9H20)

m(C8H18)=0.640-m(C9H20)

0.904=1.42*(0.640-m(C9H20))+1.405*m(C9H20)

0.9088-0.904=0.015*m(C9H20)

m(C9H20)=0.32 kg

w(C9H20)=m(C9H20)/m(smese)

w(C9H20)=0.32/0.64=0.5=50 %

w(C8H18)=100-w(C9H20)=100-50=50 %

 

 

 

Provincijalka's picture

Hvala mnogo!