Câu 4: \(n_{CO_2}=0.1mol\)

a. \(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2+H_2O\)

       0.1                                         \(\leftarrow\)  0.1

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

b. Ta có: \(0.1\times106+m_{CuO}=20.6\Leftrightarrow m_{CuO}=10g\)

\(\%m_{Na_2CO_3}=\dfrac{\left(0.1\times106\times100\right)}{20.6}=51.46\%\)

\(\%m_{CuSO_4}=100-51.46=48.54\%\)

c. \(n_{CuO}=\dfrac{10}{80}=0.125mol\)

\(\Rightarrow n_{H_2SO_4}=0.1+0.125=0.225mol\)

\(m_{dd_{H_2SO_4}}=\dfrac{\left(0.225\times98\times100\right)}{19.6}=112.5g\)

d. \(m_{dd}\) sau phản ứng \(=20.6+112.5-0.1\times44=28.7g\)

\(C\%_{Na_2SO_4}=\dfrac{\left(0.1\times142\times100\right)}{128.7}=11.03\%\)

\(C\%_{CuSO_4}=\dfrac{\left(0.125\times160\times100\right)}{128.7}=15.54\%\)