$n_{CuCl_2} = 0,2.0,1 = 0,02(mol)$
$CuCl_2 + 2NaOH \to Cu(OH)_2 + 2NaCl$
$n_{Cu(OH)_2} = n_{CuCl_2} = 0,02(mol)$
$m_{Cu(OH)_2} =0,02.98 = 1,96(gam)$
Đáp án A
Ta có: \(n_{CuCl_2}=0,1.200:1000=0,02\left(mol\right)\)
PTHH: \(CuCl_2+2NaOH--->Cu\left(OH\right)_2\downarrow+2NaCl\)
Theo PT: \(n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,02\left(mol\right)\)
=> \(m_{Cu\left(OH\right)_2}=0,02.98=1,96\left(g\right)\)
Chọn A