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 !!!!!