Question

520 grame zinc de puritate 75 %,reactioneaza cu o solutie de acid clorhidric de concentratie 36,5 %.Calculeaza a) masa solutiei de acid clorhidric utilizat.b)Numarul de moli de gaz rezultat.c)Cantitatea de sare obtinuta.

Answer 1

p=m pur/m impur*100 => m pur = p*m impur/100=75*520/100=390g Zn pur

Zn + 2HCl -> ZnCl2 + H2

a) MZn=65g/mol

MHCl=36,5g/mol

65g Zn.........................2*36,5 HCl
390g Zn.........................xg HCl

x=73*390/65=438g HCl (md)

C=md/ms*100=>ms=md*100/C=438*100/36,5= Calculeaza !!

b) 65g Zn...........................1 mol H2
390g Zn.............................x moli H2

x=390/65= Calculeaza !!

c) MZnCl2=AZn+2ACl=65+2*35,5=136g/mol

65g Zn...................136g ZnCl2
390g Zn....................xg ZnCl2

x=390*136/65=Calculeaza !!