Cho 1,792 lít O2 tác dụng hết với hỗn hợp

* Hướng dẫn giải

Chọn đáp án C

$X\left\{\begin{array}{l}Na\\ K\\ Ba\end{array}\right.\underset{}{\overset{\begin{array}{l}+{\text{ O}}_2\\ \mathrm{0,08}\text{ mol}\end{array}}{\to }}Y\left\{\begin{array}{l}Oxit\\ Kim loai dư\end{array}\right.\underset{}{\overset{+{\text{ H}}_{2}O}{\to }}Z\left\{\begin{array}{l}N{a}^{+}\\ {\text{K}}^{+}\\ {\text{Ba}}^{2+}\\ O{H}^{-}\\ \underset{\mathrm{0,14}\text{mol}}{\underbrace{H_2}}\end{array}\right.+\left|\begin{array}{l}\underset{}{\overset{+{\text{ NaHCO}}_3\text{d­}}{\to }}\underset{\mathrm{0,2}mol}{\underbrace{BaC{O}_3\downarrow }}\\ \underset{}{\overset{\begin{array}{l}+{\text{ CO}}_2\\ \mathrm{0,45}mol\end{array}}{\to }}\underset{m\text{ gam}}{\underbrace{{\text{BaCO}}_3\downarrow }}\end{array}\right.$

$\begin{array}{l}{n}_{O{H}^{-}}\stackrel{BTĐT}{=}{n}_{N{a}^{+}}+n_{K^{+}}+n_{B{a}^{2+}}\stackrel{BT\text{}E (đâu \to cuôi)}{=}4n_{O_2}+2n_{H_2}=4\frac{\mathrm{1,792}}{\mathrm{22,4}}+2\frac{\mathrm{3,136}}{\mathrm{22,4}}=\mathrm{0,6}\text{ mol}\\ {\text{n}}_{B{a}^{2+}}{\text{= n}}_{BaC{O}_3}=\frac{\mathrm{39,4}}{197}=\mathrm{0,2}{\text{ mol; n}}_{C{O}_3^{2-}}=n_{O{H}^{-}}-n_{C{O}_2}=\mathrm{0,6}-\mathrm{0,45}=\mathrm{0,15}\\ \to {\text{ n}}_{{\text{BaCO}}_3}{\text{tinh theo mol CO}}_3^{2-}\to \text{ m = 0,15.197=29,55\hspace{0.33em}gam }\end{array}$