Question

214,5 g strujitura de Zn cu 10% impuritati sunt tratate cu HCl . Calculati volumul de gaz degajat si nr de moli de sare formati .

Dau coroana

Answer 1

-calculez masa de Zn pur din p=90%

p= m,px100/m,i---> mp=214,5x90/100g=193,05g Zn

-calculez moli Zn

niu=m/M= 193,05g/65 g/mol= 2,97mol Zn

-calculez moli si vol.H2 , mol si masa de sare, din ecuatia chimica

1mol..................1mol..............1mol..

Zn + 2HCl----> ZnCl2   +    H2

2,97mol...........x=2,97mol...y=2,97mol

V,H2= niuxVm= 2,97molx22,4 l/mol= .......................calculeaza !!!!

m,ZnCl2= niuxM= 2,97gx136g/mol=.......................calculeaza

Answer 2

     193,05g       yg        xg

     Zn +2HCl=ZnCl2 +H2

     65g            136g      2g

MZnCl2=65+2.35,5=136------>1mol=136 g

se afla puritatea Zn------->p=90 la suta     [  100-10=90   ]

p=mp.100: mt

mp=p.mt :100

mp=90.214,5:100=193,05 g Zn pur

x=193,05 .2 : 65=5,94 g hidrogen

nr. moli=n

n=m: M

n=5,94 g: 2g/moli=2,97 moli hidrogen

VH2=2,97 moli .22,4 litri/moli=66,528 litri

y=193,05 .136 :65=403,92 g ZnCl2

n=403,92g :136g/moli=2,97 moli ZnCl2