Đáp án:
\(\begin{array}{l}
b)\\
{m_{NaOH}} = 40g\\
{m_{N{a_2}C{O_3}}} = 53g\\
c)\\
{V_{C{O_2}}} = 0,56l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
C{O_2} + 2NaOH \to N{a_2}C{O_3} + {H_2}O\\
b)\\
{n_{C{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{11,2}}{{22,4}} = 0,5mol\\
{n_{NaOH}} = 2{n_{C{O_2}}} = 1mol\\
{m_{NaOH}} = n \times M = 1 \times 40 = 40g\\
{n_{N{a_2}C{O_3}}} = {n_{C{O_2}}} = 0,5mol\\
{m_{N{a_2}C{O_3}}} = n \times M = 0,5 \times 106 = 53g\\
c)\\
{n_{N{a_2}C{O_3}}} = \dfrac{m}{M} = \dfrac{{2,65}}{{106}} = 0,025mol\\
{n_{C{O_2}}} = {n_{N{a_2}C{O_3}}} = 0,025mol\\
{V_{C{O_2}}} = n \times 22,4 = 0,025 \times 22,4 = 0,56l
\end{array}\)