Răspuns :
112g xg yg
Al Cl3 +3KOH =Al(OH)3 +3KCl
3.56g 78g 3.74,5g
MKOH=39+1+16=56------>1mol=56g
MAl(OH)3 =27+3.16 +3=78------->1mol=78g
MKCl=39+35,5=74,5 ------->1mol=74,5g
se afla x din ec.
x=112.78:3 .56=52 g Al(OH)3
Se afla y
y=112.3 .74,5 :3.56=149g KCl
se afla nr. de moli KCl
n=m:M
n=149:74,5=2moli KCl