Hòa tan 11,25 gam hỗn hợp Na, K, Na2O, K2O vào nước dư

* Hướng dẫn giải

Đáp án A

$\begin{array}{l}+ Sơ đồ phản ứng:\\ \underset{11,25 gam}{\underbrace{\left\{\begin{array}{l}Na: x mol\\ K: y mol\\ O: z mol\end{array}\right\}\stackrel{quy đổi}{\leftarrow }}}\underset{11,25 gam}{\underbrace{\left\{\begin{array}{l}Na, N{a}_{2}O\\ K, K_{2}O\end{array}\right\}\stackrel{H_{2}O dư}{\to }}}\underset{dd X}{\underbrace{\left\{\begin{array}{l}NaOH: 0,2 mol\\ KOH: y mol\end{array}\right\}}}+\underset{0,125 mol}{\underbrace{H_2\uparrow }}\\ + \left\{\begin{array}{l}{n}_{Na}=n_{NaOH}=x=0,2\\ m_{hỗn hợp}=23x+39y+16z=11,25\\ BTE: x+y=2z+0,125.2\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=0,2\\ y=0,15\\ z=0,05\end{array}\right.\\ + \underset{dd X}{\underbrace{\left\{\begin{array}{l}NaOH: 0,2\\ KOH: 0,15\end{array}\right\}}}\stackrel{C{O}_2}{\to }\underset{dd Y}{\underbrace{\left\{\begin{array}{l}N{a}^{+}: 0,2\\ K^{+}: 0,15\\ C{O_3}^{2-}: a\\ HC{O_3}^{-}: b\end{array}\right\}\stackrel{0,28 mol HCl (cho từ từ)}{\to }}}\underset{0,2 mol}{\underbrace{C{O}_2\uparrow }}\\ + \left\{\begin{array}{l}CT: a=n_{H^{+}}-n_{C{O}_2}=0,08\\ BTĐT: b=0,2+0,15-0,08.2=0,19\end{array}\right.\Rightarrow \left\{\begin{array}{l}BTNT C: n_{C{O}_2 p¨o}=0,27\\ V_{C{O}_2}=6,048 lít\end{array}\right.\end{array}$