PTHH:

Phần 1:2CO+O_2\rightarrow2CO_22CO+O2​→2CO2​ (1)

2H_2+O_2\rightarrow2H_2O2H2​+O2​→2H2​O (2)

CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2OCO2​+Ca(OH)2​→CaCO3​↓+H2​O (3)

Phần 2: CuO+H_2\rightarrow Cu+H_2OCuO+H2​→Cu+H2​O (4)

CuO+CO\rightarrow Cu+CO_2CuO+CO→Cu+CO2​ (5)

a, Ta có: n_{CaCO3}=\dfrac{20}{100}=0,2\left(mol\right)nCaCO3​=10020​=0,2(mol)

Theo PTHH(3): n_{CO2\left(3\right)}=n_{CaCO3}=0,2\left(mol\right)nCO2(3)​=nCaCO3​=0,2(mol)

Theo PTHH(1): n_{CO2\left(1\right)}=n_{CO2\left(3\right)}=0,2\left(mol\right)nCO2(1)​=nCO2(3)​=0,2(mol)

Vì 2 phần bằng nhau nên:

n_{CO\left(1\right)}=n_{CO\left(5\right)}=0,2\left(mol\right)nCO(1)​=nCO(5)​=0,2(mol)

\Rightarrow⇒Tổng n_{CO}=0,2\cdot2=0,4\left(mol\right)nCO​=0,2⋅2=0,4(mol)

Theo PTHH(5):n_{Cu}=n_{CO\left(5\right)}=0,2\left(mol\right)nCu​=nCO(5)​=0,2(mol)

\Rightarrow m_{Cu\left(5\right)}=0,2\cdot64=12,8\left(g\right)⇒mCu(5)​=0,2⋅64=12,8(g)

\Rightarrow m_{Cu\left(4\right)}=19,2-12,8=6,4\left(g\right)⇒mCu(4)​=19,2−12,8=6,4(g)

\Rightarrow n_{Cu\left(4\right)}=\dfrac{6,4}{64}=0,1\left(mol\right)⇒nCu(4)​=646,4​=0,1(mol)

Theo PTHH(4): n_{H2}=n_{Cu}=0,1\left(mol\right)nH2​=nCu​=0,1(mol)

\Rightarrow⇒Tổng n_{H2}=0,1\cdot2=0,2\left(mol\right)nH2​=0,1⋅2=0,2(mol)

\Rightarrow n_{hh}=0,4+0,2=0,6\left(mol\right)⇒nhh​=0,4+0,2=0,6(mol)

\Rightarrow V_{hh}=0,6\cdot22,4=13,44\left(l\right)⇒Vhh​=0,6⋅22,4=13,44(l)

b, Vì tỉ lệ về thể tích cũng là tỉ lệ về số mol nên:

\Rightarrow\%V_{CO}=\dfrac{0,2}{0,6}\cdot100\%\approx33,33\%⇒%VCO​=0,60,2​⋅100%≈33,33% \%V_{H2}=100\%-33,33\%=66,67\%%VH2​=100%−33,33%=66,67%

\Rightarrow\%m_{CO}=\dfrac{0,2\cdot28}{0,2\cdot2+0,4\cdot28}\cdot100\%\approx96,55\%⇒%mCO​=0,2⋅2+0,4⋅280,2⋅28​⋅100%≈96,55%

\%m_{H2}=100\%-96,55\%=3,45\%%mH2​=100%−96,55%=3,45%