Ta có \(n_{CuCl_2}=\frac{1,35}{135}=0,01\left(mol\right);n_{KOH}=\frac{2,8}{56}=0,05\left(mol\right)\)
PTHH \(CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2\downarrow+2KCl\) (1)
Thấy \(n_{CuCl_2pu}:n_{KOHpu}=\frac{0,01}{1}< \frac{0,05}{2}\Rightarrow CuCl_2hết\)tính theo \(CuCl_2\)
a) \(Cu\left(OH\right)_2-t^o\rightarrow CuO+H_2O\) (2)
\(\Rightarrow\) Chất rắn sau khi nung là CuO
Theo (1) : \(n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,01\left(mol\right)\)
Theo (2) : \(n_{CuO}=n_{Cu\left(OH\right)_2}=0,01\left(mol\right)\)
\(\Rightarrow m_{CuO}=80\times0,01=0,8\left(g\right)\)
b) Chất có trong dd sau phản ứng là KOH
Theo (1) \(n_{KOHpu}=2n_{CuCl_2}=0,02\left(mol\right)\)
\(\Rightarrow n_{KOHdu}=0,05-0,02=0,03\left(mol\right)\)
\(\Rightarrow m_{KOHdu}=0,03\times56=1,68\left(g\right)\)