Hấp thụ hoàn toàn 4,48 lít khí CO2(đkc) vào 300ml dd NaOH 2M thu được dung dịch X. a) Tính khối lượng muối tan trong dd X. b)Tính nồng độ các chất trong dd X

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Đáp án:

\(\begin{array}{l}
a)\\
{m_{N{a_2}C{O_3}}} = 21,2g\\
b)\\
{C_M}NaOH = {C_M}N{a_2}C{O_3} = 0,67M
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
{n_{NaOH}} = 0,3 \times 2 = 0,6\,mol\\
{n_{C{O_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = \dfrac{{0,6}}{{0,2}} = 3\\
T > 3 \Rightarrow \text{ Tạo muối $Na_2CO_3$ }\\
2NaOH + C{O_2} \to N{a_2}C{O_3} + {H_2}O\\
{n_{N{a_2}C{O_3}}} = {n_{C{O_2}}} = 0,2\,mol\\
{m_{N{a_2}C{O_3}}} = 0,2 \times 106 = 21,2g\\
b)\\
{n_{NaOH}}  \text{ dư }= 0,6 - 0,2 \times 2 = 0,2\,mol\\
{C_M}NaOH \text{ dư }= = {C_M}N{a_2}C{O_3} = \dfrac{{0,2}}{{0,3}} = 0,67M
\end{array}\)

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