\(n_{CuCl_2}=\dfrac{135}{135}=1\left(mol\right);n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)
\(PTHH:CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaCl\)
Sau p/ứ không có chất dư.
=> Kết tủa là Cu(OH)2,dd sau p/ứ là NaCl
a_)\Theo PTHH: \(n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Cu\left(OH\right)_2}=0,1.98=9,8\left(g\right)\)
b_)Theo PTHH\(n_{NaCl}=n_{NaOH}=0,5\left(mol\right)\)
\(\Rightarrow m_{NaCl}=0,5.58,5=29,25\left(g\right)\)