Question

pronlema incercuita repede

Answer 1

Notez inmultirile cu * si fractiile cu /  
Zn+2HCl--> ZnCl2+H2
MZn=65
1 mol Zn= 65g 
MHCl = 1*35,5=36,5 
1 mol HCl = 36,5 g 
c=md*100/ms=>md=c*ms/100=25*36,5/100=9,125 HCl  
In 65 g Zn..... 2*36,5 HCl 
In  a g Zn ...... 9,125 HCl 

65/a=2*36,5/9,125 => a= 65*9,125/2*36,5= 8,125 g Zn 
Daca in reactie a ramas 10g de Zn atunci masa necunoscuta (X) de Zn introdus este 
x = 10+a = 10+8,125= 18,125 g Zn