Question

Un cui din Fe de 6g are 5% impuritati.Acesta se introduce in solutie HCl 5%. Se cere : a) cantitatea de Fe pura ; b) ecuatia chimica ; c) masa solutiei de HCl consumata ;d) volumul de hidrogen degajat.

Answer 1

m Fe impur = 6 g 
p Fe = 95%
c HCl = 5%

a) p = mp / mi x100 => mp = mi x p / 100 
mp Fe = 6 x 95 / 100 = 5.7 g Fe pur
p=puritatea
mp = masa pura
mi = masa impura

b) 2 Fe + 6 HCl = 2 FeCl3 + 3 H2

c) 
2 x 55.84 g/mol Fe ..... 6 x 36.5 g/mol HCl
              5.7 g Fe ............a g HCl
a= 5.7 x 6 x 36.5 / 2 x 55.84 = 11.18 g HCl (md)
c= md / ms x 100 => ms = md x 100/ c
ms = 11.18 x 100 / 5 = 223.55 g sol. HCl 5%
c= concentratia
md=masa de dizolvat
ms= masa solutie

d)
2 x 55.84 g/mol Fe .... 3 x 2g/mol
5.7 g Fe ......................c g H2
c= 5.7 x 3 x 2 / 2 x  55.84  = 0.31 g H2
n H2 = m H2 / M H2 =0.31 g / 2 g/mol = 0.155 mol H2
V = n x 22.4 l/mol => V H2 = 0.155 mol x 22.4 l/mol =  3.47 l H2 (cn)
n= nr de moli
V = volumul
 cn = conditii normale (1 atm, 0 grade C)