Răspuns:
Explicație:
350g sol. CaCl2 , c=20%
200g sol. AgNO3 ,c= 20%
subst . in exces
masa AgCl
-se afla md sol. de CaCl2
md= c . ms : 100
md= 20 . 350 : 100= 70g CaCl2
-se afla md sol. de AgNO3
md= 200 . 20 :100= 40g AgNO3
-se presupune ca nu se cunoaste masa de CaCl2
xg 40g yg
CaCl2 + 2Ag NO3= Ca(NO3)2 + 2AgCl
111 g 2.170g 2.133,5g
x= 111 . 40 : 340= 13,06 gCaCl2
Deci, CaCl2 se afla in exces-------> 70 - 13, 06=56,94 g CaCl2 in exces
y= 40 . 2 . 133,5 : 340= 31,41 g AgCl
MCaCl2= 40 + 2.35,5=111----> 111 g/moli
MAgNO3= 108 + 14 + 3.16=170-----> 170g/moli
mAgCl= 108 + 35,5= 133,5------> 133,5g/moli
