Răspuns:
Explicație:
1.
FeCl2
r.a.=1 :2
r.m.= 56 : 71
comp. proc. prin masa moleculara
M= 56 + 71=127-------> 127g/moli
127g---------56gFe---------71gclor
100g-----------x------------------y
x=100.56 : 127=44,09%Fe
y=100.71 : 127=55,91% clor
2.
Fe + CuSO4=FeSO4 + Cu
Mg + K2SO4 -----> nu are loc
Ag + H2O------> nu are loc
Ca + 2HCl= CaCl2 +H2
3.
--se afla masa pura de Cu
p=80% [100-20=80 ]
mp=p.mi : 100
mp=80 . 16 :100=12,8 g Cu
12,8g yg xg
2Cu + O2 =2CuO
2.64g 32g 2.80g
x=12,8 . 160 : 128=16g CuO
n= 16g : 160g/moli=0,1moli Cu
y=12,8 . 32 : 128=3,2goxigen
n=3,2g : 32g/moli=0,1moli O2
MCuO=64 +16=80-------->80g/moli