a, CuCl2 + 2NaOH -> Cu(OH)2 + 2 NaCl
b,n CuCl2 = 0.15*2=0.3 (mol)
=> n NaOH = 2*0.3=0.6 (mol)
m dd= m ct * 100% /C% = (0.6*40) * 100%/20% = 120 (g)
c, n Cu(OH)2 = 0.3 (mol)
=> m Cu(OH)2 = 0.3*98 = 29.4 (g)
m CuCl2 = 0.3*135=40.5 (g)
m NaOH = 0.6*57 = 34.2 (g)
m dd(sau phản ứng) = m CuCl2 + m Na(OH)2 - m Cu(OH)2 = 40.5 + 34.2-29.4=45.3(g)
d, Cu(OH)2 -> CuO + H2O
n CuO = n Cu(OH)2 =0.3 (mol)
=> m CuO = 0.3*80=24 (g)