Question

Se dau 7,6 g Zn ,de puritate 70% care reactioneaza cu o solutie de 9% H2SO4 .Se cere: a)masa solutiei de h2o ; b)masa de sare formata; c)concentratia finala a soutiei.

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Answer 1

Răspuns:

Zn +H2SO4-->ZnSO4+H2

puritate=(mp*100)/mip

mp=(70*7,2)/100=5,04

65gZn...........................98gH2SO4

5,04...............................x=7,59

c=9%

md=7,59g

c=(md/ms)*100

ms2=(md/c)*100=84,3 g

b)

65gZn....................................161gZnSO4

5,04............................................Y=12.4g

c)

c1=md1*100/ms1

ms1=56g

ms2=84g

msf=140g

cf=(md/msf)*100

cf=8,85%

...............................................................

Explicație: