Câu 20:
\[n_{CO_2}=\dfrac{13,2}{44}=0,3(mol)\]
\[n_{O_2}=\dfrac{12,8}{32}=0,4(mol)\]
PTHH:
\[2CO + O_2\to 2CO_2\qquad(1)\]
\[2H_2+O_2\to 2H_2O\qquad(2)\]
Theo PTHH $(1)\to n_{CO_2}=n_{CO}=0,3(mol)$
$\to m_{CO}=28.0,3=8,4(g)$
Theo PTHH $(1)\to n_{O_2(1)}=\dfrac{1}{2}.n_{CO_2}=\dfrac{1}{2}.n_{CO_2}=0,15(mol)$
$\to n_{O_2(2)}=0,4-0,15=0,25(mol)$
\[\to n_{H_2}=0,25.2=0,5(mol)\]
\[\to m_{H_2}=0,5.2=1(g)\]
\[\to \%m_{CO}=\dfrac{8,4}{8,4+1}\times 100\%=89,36\%\]
\[\to \%m_{H_2}=100\%-89,36\%=10,64\%\]