Đáp án:
\(\begin{array}{l}
V{D_3}:\\
m = 41,1g\\
V{D_4}:\\
{m_{N{a_2}C{O_3}}} = 31,8g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
V{D_3}:\\
{n_{C{O_2}}} = \dfrac{{10,08}}{{22,4}} = 0,45\,mol\\
{n_{NaOH}} = 0,5 \times 1,2 = 0,6mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = \dfrac{{0,6}}{{0,45}} = 1,333\\
1 < T < 2 \Rightarrow \text{ Tạo cả 2 muối}\\
2NaOH + C{O_2} \to N{a_2}C{O_3} + {H_2}O(1)\\
N{a_2}C{O_3} + C{O_2} + {H_2}O \to 2NaHC{O_3}(2)\\
{n_{N{a_2}C{O_3}(1)}} = {n_{C{O_2}(1)}} = \dfrac{{0,6}}{2} = 0,3\,mol\\
{n_{N{a_2}C{O_3}(2)}} = {n_{C{O_2}(2)}} = 0,45 - 0,3 = 0,15\,mol\\
{n_{N{a_2}C{O_3}}} = 0,3 - 0,15 = 0,15\,mol\\
{n_{NaHC{O_3}}} = 0,15 \times 2 = 0,3\,mol\\
m = {m_{N{a_2}C{O_3}}} + {m_{NaHC{O_3}}} = 0,15 \times 106 + 0,3 \times 84 = 41,1g\\
V{D_4}:\\
C{O_2} + 2NaOH \to N{a_2}C{O_3} + {H_2}O\\
{n_{C{O_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_{N{a_2}C{O_3}}} = {n_{C{O_2}}} = 0,3\,mol\\
{m_{N{a_2}C{O_3}}} = 0,3 \times 106 = 31,8g
\end{array}\)
