\(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)
\(n_{HCl}=0,4.1=0,4\left(mol\right)\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
0,15 ---> 0,3 -----> 0,15 ----> 0,15
Xét: \(\dfrac{0,15}{1}< \dfrac{0,4}{2}\) => axit dư
a. HCl dư sau phản ứng và \(n_{HCl.dư}=0,4-0,3=0,1\left(mol\right)\)
b.
Trong dung dịch A có:
\(m_{HCl}=0,1.36,5=3,65\left(g\right)\)
\(m_{CuCl_2}=0,15.135=20,25\left(g\right)\\ m_{H_2O}=0,15.18=2,7\left(g\right)\)
\(a)n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\\ n_{HCl}=0,4.1=0,4\left(mol\right)\\ CuO+2HCl\xrightarrow[]{}CuCl_2+H_2O\\ \Rightarrow\dfrac{0,15}{1}< \dfrac{0,4}{2}\Rightarrow HCl.dư\\n_{HCl\left(pư\right)}=0,15.2=0,3\left(mol\right)\\ n_{HCl\left(dư\right)}=0,4-0,3=0,1\left(mol\right)\\ b) n_{CuCl_2}=n_{CuO}=n_{H_2O}=0,15mol\\ m_{CuCl_2}=0,15.135=20,25\left(g\right)\\ m_{H_2O}=0,15.18=2,7\left(g\right)\\ m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\)
