Đáp án:

 b)

\({m_{HC{l_{tgpu}}}} = 3,65g\)

c)

\({V_{C{O_2}}} = 1,12l\)

d)

\(C{\% _{N{a_2}C{O_3}}} = 2,65\% \)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
N{a_2}C{O_3} + 2HCl \to 2NaCl + C{O_2} + {H_2}O\\
b)\\
{m_{HC{l_{tgpu}}}} = m \times 3,65\%  = 100 \times 3,65\%  = 3,65g\\
c)\\
{n_{HCl}} = \dfrac{m}{M} = \dfrac{{3,65}}{{36,5}} = 0,1mol\\
{n_{C{O_2}}} = \dfrac{1}{2} \times {n_{HCl}} = \dfrac{1}{2} \times 0,1 = 0,05mol\\
{V_{C{O_2}}} = n \times 22,4 = 0,05 \times 22,4 = 1,12l\\
d)\\
{n_{N{a_2}C{O_3}}} = {n_{C{O_2}}} = 0,05mol\\
{m_{N{a_2}C{O_3}}} = n \times M = 0,05 \times 106 = 5,3g\\
C{\% _{N{a_2}C{O_3}}} = \dfrac{{{m_{ct}}}}{{{m_{{\rm{dd}}}}}} \times 100\%  = \dfrac{{5,3}}{{200}} \times 100\%  = 2,65\% 
\end{array}\)