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Răspuns :

Puia68
md=4g
m,apa=16g
ms=md+m,apa=20g
b)  c= mdx100/ms==400/20---> 20%
a) 2NaOH +FeSO4__>Na2SO4+Fe(OH)2
   2x40g..........151g
     2g..............m=3,77gFeSO4--MASA NECESARA

7,6/100= md/200-->md=15,2g FeSO4 masa disponibila este  in esces
exces sare:m=15,2-3,77-= >11,43g
200g sol...........15,2g sare
ms......................11,43-----> ms, exces= 146g

6NaOH+Al2(SO4)3---> 3Na2SO4 +2aL(OH)3↓
6X40g  ..........................................      2x78g
 2g..........................................................m= 1,3gAl(OH)3

d) cred ca se cere c% in care md=m,Na2SO4  =3,55G  
                                                    m,s= ms,exces+m apa solNaOH  =146+8
 C/100= 3,55X100/ 154     nu sunt sigura !!!!!