Răspuns:
-calculez masa de zinc pur
p=mpx100/m,i---> mp=1300x50/100g= 650g Zn pur
-calculez masa de acid din solutie
c=mdx100/ms---> md= 36,5x2500/100g= 912,5gHCl
-calculez moli Zn si HCl
niu=m/A,Zn= 650/65mol=10mol Zn
niu=m/.miu,HCl=912,5/36,5mol=25mol HCl
-deduc substanta in exces din ecuatia chimica
1mol.......2mol........................1mol
Zn + 2HCl-----> ZnCl2 + H2
10mol.....x......................................y
x= 20molHCl; dar sunt 25mol, deci 5mol sunt in exces
y=10molH2
-V=niuxVm= 10x22,4 l=224 lH2
Explicație: