Question

Un amestec de metanol si etanol are un conținut de 12,726% H( procente de masa).Sa se calculeze cat metanol conține amestecul.

Answer 1

presupunem

a-g metanol

b-g etanol

si ca a+b=100

 

32g.......................4

a..............................y

y=4xa/32

46g........................6Gh

b...........................z

z=6xb/46

b=100-a

100..........................4xa/32+6xb/46

100...........................12,726

100x12,726=100(4xa/32+6xb/46) se simplifica prin 100

12,726=4xa/32+6xb/56

Se inlocuieste b

12,726=4xa/32+6(100-a)/46

Se aduce la acelasi numitor

12,726x32x46=4xax46+6x32(100-a)

18732,672=184xa+19200-192xa

 

              8xa=467,328

a=467,328/8

a=58,416

Answer 2

32x+46y.................4x+6y g H
100g........................12,726 g H

12,726(32x+46y) = 400x+600y
7,232x = 14,604y

x/y = 14,604/7,232 => x = 2,0193y

110,619y = m

%alcool metilic = 64,619/110,619×100 = 58,4

=> m alcool metilic = 58,4g

Calculul se face cu toate zecimalele