${n_{{K_2}C{O_3}}} = {n_{NaHC{O_3}}} = a\,mol;\,{n_{Ba{{(HC{O_3})}_2}}} = b\,mol \to \left\{ \begin{gathered}{n_{CO_3^{2 - }}} = a\,mol \hfill \\{n_{HCO_3^ - }} = (a + 2b)\,mol \hfill \\ \end{gathered} \right.$

$2{H^ + } + CO_3^{2 - }\xrightarrow{{}}C{O_2} \uparrow + {H_2}O$

${H^ + } + HCO_3^ - \xrightarrow{{}}C{O_2} \uparrow + {H_2}O$

${n_{HCl}} = 0,5.0,56 = 0,28\,mol \to 2a + (a + 2b) = 0,28\,\,\,(1)$

$HCO_3^ - + O{H^ - }\xrightarrow{{}}CO_3^{2 - } + {H_2}O$

$\,{n_{NaOH}} = 0,2.1 = 0,2\,mol \to a + 2b = 0,2\,\,\,(2)$

$ (1),(2) \to a = 0,04;\,\,b = 0,08$

${K_2}C{O_3} + Ba{(HC{O_3})_2}\xrightarrow{{}}BaC{O_3} \downarrow + 2KHC{O_3}$

0,04 0,08 → 0,04

${m_{BaC{O_3}}} = 0,04.197 = 7,88\,g$