Đáp án:
$a = 0,04$
Giải thích các bước giải:
${n_{C{O_2}}} = \frac{{2,688}}{{22,4}} = 0,12\,\,mol;\,\,{n_{Ba{{(OH)}_2}}} = 2,5a\,\,mol$
${n_{BaC{O_3}}} = \frac{{15,76}}{{197}} = 0,08\,\,mol$
Nhận thấy ${n_{BaC{O_3}}} < {n_{C{O_2}}} \to $ Phản ứng tạo hai muối $BaC{O_3};\,\,Ba{(HC{O_3})_2}$
Phương trình hóa học:
$\begin{gathered} 2C{O_2} + Ba{(OH)_2} \to Ba{(HC{O_3})_2} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, 0,04 \,\,\,\,\,\,\,\,\,\,\,\,\to 0,02 \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,0,02\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol \hfill \\ C{O_2} + Ba{(OH)_2} \to BaC{O_3} + {H_2}O \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,08 \leftarrow \,\,\,\,\,0,08\,\,\, \leftarrow \,\,\,\,\,\,0,08\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol \hfill \\ \to {n_{Ba{{(OH)}_2}}} = 0,02 + 0,08 = 0,1 = 2,5a\,\,\,mol \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,a = 0,04 \hfill \\ \end{gathered} $
