Răspuns :
se da
m(sol)25 g
w%(HCl)=36,5%
10g de Zn nereactionat
Zn+2HCl=ZnCl2+H2
aflam masa acidului
w%(HCl)=m(HCl)*100%/m(sol)
m(HCl)=w%(HCl)*m(sol+ /100% = 36,5 *25 /100 =9,125 g
aflam niu(HCl)
niu(HCl)=m(HCl)/M(HCl)=9,125 g / 36,5 =0,25 mol
aflam cit zin o intrat in reactie shi m alui
niu(Zn)=1/2 *niu(HCl)= 0,125 mol
aflam masa zincului intrat in reactie
m(Zn)=niu(Zn)*M(Zn)=8,125 g
astfel daca au ramas 10 grame shi in reactie au intrat 8,125 g de Zn
atunci x= 18,125 g
m(sol)25 g
w%(HCl)=36,5%
10g de Zn nereactionat
Zn+2HCl=ZnCl2+H2
aflam masa acidului
w%(HCl)=m(HCl)*100%/m(sol)
m(HCl)=w%(HCl)*m(sol+ /100% = 36,5 *25 /100 =9,125 g
aflam niu(HCl)
niu(HCl)=m(HCl)/M(HCl)=9,125 g / 36,5 =0,25 mol
aflam cit zin o intrat in reactie shi m alui
niu(Zn)=1/2 *niu(HCl)= 0,125 mol
aflam masa zincului intrat in reactie
m(Zn)=niu(Zn)*M(Zn)=8,125 g
astfel daca au ramas 10 grame shi in reactie au intrat 8,125 g de Zn
atunci x= 18,125 g