Question

Salut,la problema 5 m am incurcat,banuiesc ca este un exces,la a) ma refer dar nu sunt sigura..

Answer 1

Cu(OH)2+H2SO4⇒CuSO4+2H2O
ω(Cu(OH)2)=msubs./m sol.*100%
m (Cu(OH)2=196 g*75%/100%=147g
n-cantitatea de substanta
n(Cu(OH)2)=147/98=1,5 mol
n(H2SO4)=196/94=2,08 mol
1,5 mol<2,08 ⇒ in exces H2SO4
conform ecuatii chimice
n (CuSO4)=1,5*1/1=1,5 moli
R/s:n(CuSO4)=1,5 moli