Cho 8,96 lít O2 tác dụng hết với

* Hướng dẫn giải

Chọn đáp án C

$X\left\{\begin{array}{l}Na\\ K\\ Ba\end{array}\right.\underset{}{\overset{\begin{array}{l}+{\text{ O}}_2\\ \mathrm{0,4}\text{ mol}\end{array}}{\to }}Y\left\{\begin{array}{l}Oxit\\ Kim\text{\hspace{0.33em}loai du}\end{array}\right.\underset{}{\overset{+{\text{ H}}_{2}O}{\to }}Z\left\{\begin{array}{l}N{a}^{+}\\ {\text{K}}^{+}\\ {\text{Ba}}^{2+}\\ O{H}^{-}\\ \underset{\mathrm{0,7}\text{mol}}{\underbrace{H_2}}\end{array}\right.+\left|\begin{array}{l}\underset{}{\overset{+{\text{ NaHCO}}_3\text{d­}}{\to }}\underset{1mol}{\underbrace{BaC{O}_3\downarrow }}\\ \underset{}{\overset{\begin{array}{l}+{\text{ CO}}_2\\ \mathrm{2,25}mol\end{array}}{\to }}\underset{m\text{ gam}}{\underbrace{{\text{BaCO}}_3\downarrow }}\end{array}\right.$

$n_{O{H}^{-}}\stackrel{BTÐT}{=}{n}_{N{a}^{+}}+n_{K^{+}}+n_{B{a}^{2+}}\stackrel{BT\text{}E\text{(dau}\to cuoi)}{=}4n_{O_2}+2n_{H_2}=4\frac{\mathrm{8,96}}{\mathrm{22,4}}+2\frac{\mathrm{15,68}}{\mathrm{22,4}}=3\text{ mol}$

$\begin{array}{l}{\text{n}}_{B{a}^{2+}}{\text{= n}}_{BaC{O}_3}=\frac{197}{197}=1{\text{ mol; n}}_{C{O}_3^{2-}}=n_{O{H}^{-}}-n_{C{O}_2}=3-\mathrm{2,25}=\mathrm{0,75}\text{\hspace{0.33em}mol }\\ \to {\text{ n}}_{{\text{BaCO}}_3}{\text{tinh theo mol CO}}_3^{2-}\to \text{ m = 0,75.197 = 147,75\hspace{0.33em}gam }\end{array}$