a, PTHH:
2NaOH + CuCl2 \(\rightarrow\) 2NaCl + Cu(OH)2 \(\downarrow\)
b, Đổi : 100ml = 0,1(l)
Ta có : n NaOH = 0,1 . 1 = 0,1(mol)
Theo PTHH : n Cu(OH)2 = \(\dfrac{1}{2}\) n NaOH
= \(\dfrac{1}{2}\). 0,1
= 0,05(mol)
m Cu(OH)2 = 0,05 . 98 = 4,9(g)
c, Cu(OH)2 \(\rightarrow\) CuO + H2O
Theo PTHH : n CuO = n Cu(OH)2
= 0,05(mol)
m CuO = 0,05 . 80 = 4 (g)