\(n_{CuCl2}=\frac{54}{135}=0.4mol\)
\(n_{NaOH}=\frac{40}{40}=1mol\)
PTHH: \(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\)
\(0.4\) mol : 1 mol
Lập tỉ lệ: \(\frac{0.4}{1}< \frac{1}{2}\) ( NaOH dư, tính toán dựa vào CuCl\(_2\)
\(\rightarrow n_{CuCl2}=n_{Cu\left(OH\right)2}=0.4mol\)
\(n_{NaOH}=n_{NaCl}=2n_{CuCl2}=0.4\cdot2=0.8mol\)
\(\rightarrow n_{NaOHdư}=1-0.8=0.2mol\)
a, \(m_{Cu\left(OH\right)2}=0.4\cdot98=39.2\left(g\right)\)
\(m_{NaOHdư}=0.2\cdot40=8\left(g\right)\)
b, \(m_{NaCl}=0.4\cdot58.5=23.4\left(g\right)\)