Dẫn a mol hỗn hợp X (gồm hơi nước và khí CO2) qua cacbon nung đỏ, thu được 1,8a mol hỗn hợp khí Y gồm H2, CO và CO2

* Hướng dẫn giải

$\begin{array}{l}+amol\left\{\begin{array}{l}C{O}_2\\ H_{2}O\end{array}\right\}\stackrel{C,t^o}{\to }1,8amol\left\{\begin{array}{l}C{O}_2\\ H_2\\ CO\end{array}\right\}\stackrel{(F{e}_2{O}_3,CuO),t^o}{\to }\left\{\begin{array}{l}C{O}_2\\ H_{2}O\end{array}\right\}\\ +\left\{\begin{array}{l}BCPÖ:C\stackrel{O/oxit}{\to }C{O}_2\\ BTNT:n_C=1,8a-a=0,8a\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_C=\frac{1}{2}{n}_{O/oxitbịtáchra}=\frac{1}{2}.\frac{1,28}{16}=0,04\\ 0,8a=0,04\end{array}\right.\Rightarrow \boxed{a=0,05}\end{array}$

$\begin{array}{l}+\underset{0,05mol}{\underbrace{\left\{\begin{array}{l}C{O}_2:(0,05-z)mol\\ H_{2}O:zmol\end{array}\right\}}}\underset{t^o}{\overset{C}{\to }}Ygồm\underset{0,09mol}{\underbrace{\left\{\begin{array}{l}CO:xmol\\ C{O}_2:ymol\\ H_2:zmol\end{array}\right\}}}\\ \Rightarrow \left\{\begin{array}{l}{n}_Y=x+y+z=0,09\\ n_O=2(0,05-z)+z=x+2y\end{array}\right.\Rightarrow \left\{\begin{array}{l}x+y+z=0,09\\ x+2y+z=0,1\end{array}\right.\Rightarrow y=0,01\Rightarrow n_{BaC{O}_3}=0,01\Rightarrow \boxed{m_{BaC{O}_3}=1,97gam}\end{array}$

Chọn C