\(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)

a) \(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\left(1\right)\)

     \(Cu\left(OH\right)_2\xrightarrow[t^o]{}CuO+H_2O\left(2\right)\)

b) \(Pt\left(1\right):n_{Cu\left(OH\right)2}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)

  \(Pt\left(2\right):n_{Cu\left(OH\right)2}=n_{CuO}=0,25\left(mol\right)\Rightarrow m_{Cu}=0,25.64=16\left(g\right)\)

c) Pt(1) : \(n_{NaOH}=n_{NaCl}=0,5\left(mol\right)\Rightarrow m_{NaCl}=0,5.58,5=29,25\left(g\right)\)

\(\%N=100\%-61,017\%-15,254\%=23,729\%\)

Gọi CTTQ của X : C_xH_yN_t (x,y,t nguyên dương)

\(x:y:t=\dfrac{\%C}{12}:\dfrac{\%H}{1}:\dfrac{\%N}{14}=\dfrac{61,017}{12}:\dfrac{15,254}{1}:\dfrac{23,729}{14}=3:9:1\)

\(\rightarrow CTĐGN:\left(C_3H_9N\right)\)

\(d_{\dfrac{X}{H2}}=\dfrac{M_X}{M_{H2}}=29,5\Rightarrow M_X=29,5.2=59\)

\(\Rightarrow\left(12.3+9+14\right).n=59\rightarrow n=1\)

\(\Rightarrow CTPT:C_3H_9N\)

\(CH_3-CH_2-CH_2-NH_2\)

\(CH_3-CH\left(NH_2\right)-CH_3\)

\(CH_3-CH_2-NH-CH_3\)