Explicație:
a) lScriem ecuația reactiei :
3KOH + AlCl3 = Al(OH)3 + 3KCl
mK =39
mO=16
mH=1
mAl=27
mCl=35,5
Calculam
MKOH = mK+ mO+mH =39+16+1=56g/mol
MAl(OH)3= =mAl+3(mO+mH)=27+3(16+1)=27+3×17=27+51=78 g/mol
Conform reactiei avem:
3 KOH deci 3×MKOH =3×56= 168
Și aplicam regula de 3 simple:
168g KOH............. 78g Al(OH)3
84gKOH...............y => y= 84×78/168=>
=>y = 39g Al(OH)3
1 mol......... 78gAl(OH)3
x moli....... 39gAl(OH)3 => x= 39×1/78=> x=0,5 moli
b) C% =md/ms×100;md=133,5, C%=10
MAlCl3 = mAl+3×mCl=27+3×35,5=27+106,5=133,5
=> 10=133,5/ms ×100=>ms=133,5/10×100 =>ms=1335g AlCl3
MKCl= mK +mCl =39+35,5=74,5
Cf.reactiei 3KCl=> 3×74,5= 223,5
C%(KCl) =md(KCl)/ms(KCl) ×100
C% =74,5/223,5×100=33,(33)%
