Răspuns:
date necesare M,HCl= 36,5g/mol Vm= 22,4l/mol
-calculez moliH2
niu=V/Vm= 112 l/22,4 l/mol= 5molH2
-deduc moli Zn,HCl,ZnCl2, din ecuatia chimica
1mol......2mol...........1mol.........1mol
Zn + 2HCl=====> ZnCl2 + H2
x...........y....................z.................5mol
c)z= 5mol ZnCl2
a)m,Zn= niuxM= 5molx65g/mol=325g este m,pur
p= mpx100/m,i= 325x100/600%=54%
b) m,HCl= niuxM= 10molx36,5g/mol= 365g.......md
c= mdx100/ms---> ms= 365x100/36,5g= 1000g
Explicație:
