Dẫn từ từ 5,6 lít khí CO2 (đktc) vào 400 ml dung dịch chứa, đồng thời

* Hướng dẫn giải

Đáp án A

$\underset{0,25 \mathrm{mol}}{\underbrace{C{O}_2}}+\left\{\begin{array}{l}\underset{0,12 mol}{\underbrace{NaOH}}\\ \underset{0,08\mathrm{mol}}{\underbrace{KOH}}\\ \underset{0,075 \mathrm{mol}}{\underbrace{{\mathrm{Na}}_2{\mathrm{CO}}_3}}\\ \underset{0,05 \mathrm{mol}}{\underbrace{K_{2}C{O}_3}}\end{array}\right.\to X\left\{\begin{array}{l}\begin{array}{c}{\mathrm{Na}}^{+}:\stackrel{\mathrm{BTNT}.\mathrm{Na}}{\to }{\mathrm{n}}_{{\mathrm{Na}}^{+}}=0,27\\ {\mathrm{K}}^{+}:\stackrel{BTNT.K}{\to }{n}_{K^{+}}=0,18\\ {\mathrm{CO}}_3^{2-}:\mathrm{amol}\end{array}\\ {\mathrm{HCO}}_3^{-}:\mathrm{bmol}\end{array}\right.\stackrel{+CaC{l}_{2}d-}{\to }\underset{amol}{\underbrace{CaC{O}_3}}$

$\to \left\{\begin{array}{l}\stackrel{BTNT.C}{\to }a + b = 0,25 + 0,075 + 0,05\\ \stackrel{BTNT cho X}{\to }2a + b = 0,27 + 0,18\end{array}\right.\to \left\{\begin{array}{l}a=0,075\\ b=0,3\end{array}\right.$

$\stackrel{BTTNT.C}{\to }{n}_{CaC{O}_3}=n_{C{O}_3^{2-}}=0,075\to m_{CaC{O}_3}=0,075.100=7,5 \mathrm{gam}$