Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{n_{C{O_2}}} = 0,1mol\\
{m_{Ba{{(OH)}_2}}} = \dfrac{{150 \times 17,1\% }}{{100\% }} = 25,65g\\
\to {n_{Ba{{(OH)}_2}}} = 0,15mol\\
\to \dfrac{{{n_{Ba{{(OH)}_2}}}}}{{{n_{C{O_2}}}}} = \dfrac{{0,15}}{{0,1}} = 1,5
\end{array}\)
=> Tạo 2 muối: \(BaC{O_3}\) và \(Ba{(HC{O_3})_2}\)
Gọi a và b là số mol của CO2 (1) và CO2 (2)
\(\begin{array}{l}
C{O_2} + Ba{(OH)_2} \to BaC{O_3} + {H_2}O(1)\\
2C{O_2} + Ba{(OH)_2} \to Ba{(HC{O_3})_2}(2)\\
\left\{ \begin{array}{l}
a + b = 0,1\\
a + \frac{1}{2}b = 0,15
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,2\\
b = - 0,1
\end{array} \right.\\
b)\\
{n_{C{O_2}}} = 0,1mol\\
{m_{Ba{{(OH)}_2}}} = \dfrac{{150 \times 17,1\% }}{{100\% }} = 17,1g\\
\to {n_{Ba{{(OH)}_2}}} = 0,1mol\\
\to \dfrac{{{n_{Ba{{(OH)}_2}}}}}{{{n_{C{O_2}}}}} = \dfrac{{0,1}}{{0,1}} = 1
\end{array}\)
=> Tạo 1 muối trung hòa: \(BaC{O_3}\)
\(\begin{array}{l}
C{O_2} + Ba{(OH)_2} \to BaC{O_3} + {H_2}O\\
{n_{BaC{O_3}}} = {n_{C{O_2}}} = 0,1mol\\
\to {m_{BaC{O_3}}} = 19,7g\\
{m_{dd}} = {m_{C{O_2}}} + {m_{ddBa{{(OH)}_2}}} = 0,1 \times 44 + 100 = 104,4g\\
\to C{\% _{BaC{O_3}}} = \dfrac{{19,7}}{{104,4}} \times 100\% = 18,87\% \\
c)\\
{n_{C{O_2}}} = 0,1mol\\
{m_{Ba{{(OH)}_2}}} = \dfrac{{150 \times 17,1\% }}{{100\% }} = 128,25g\\
\to {n_{Ba{{(OH)}_2}}} = 0,75mol\\
\to \dfrac{{{n_{Ba{{(OH)}_2}}}}}{{{n_{C{O_2}}}}} = \dfrac{{0,75}}{{0,1}} = 7,5
\end{array}\)
=> Tạo 1 muối axit: \(Ba{(HC{O_3})_2}\)
\(\begin{array}{l}
2C{O_2} + Ba{(OH)_2} \to Ba{(HC{O_3})_2}\\
{n_{Ba{{(HC{O_3})}_2}}} = \dfrac{1}{2}{n_{C{O_2}}} = 0,1mol\\
\to {m_{Ba{{(HC{O_3})}_2}}} = 25,9g\\
\to {m_{dd}} = {m_{C{O_2}}} + {m_{ddBa{{(OH)}_2}}} = 0,1 \times 44 + 200 = 204,4g\\
\to C{\% _{Ba{{(HC{O_3})}_2}}} = \dfrac{{25,9}}{{204,4}} \times 100\% = 12,67\% \\
\end{array}\)
Bạn xem lại đề câu a nha.
