Đáp án:
\(\begin{array}{l}
b)\\
C{M_{HCl(dư)}} = C{M_{NaCl}} = 0,33M\\
c)\\
{m_{NaCl}} = 5,85g\\
d)\\
C{\% _{NaCl}} = 2,03\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)N{a_2}C{O_3} + 2HCl \to 2NaCl + C{O_2} + {H_2}O\\
b)\\
{n_{N{a_2}C{O_3}}} = 0,1mol\\
{n_{HCl}} = 0,3mol\\
\to \dfrac{{{n_{N{a_2}C{O_3}}}}}{1} < \dfrac{{{n_{HCl}}}}{2} \to {n_{HCl}}dư\\
\to {n_{NaCl}} = {n_{N{a_2}C{O_3}}} = 0,1mol\\
\to {n_{HCl}} = 2{n_{N{a_2}C{O_3}}} = 0,2mol\\
\to {n_{HCl(dư)}} = 0,1mol\\
\to C{M_{HCl(dư)}} = C{M_{NaCl}} = \dfrac{{0,1}}{{0,1 + 0,2}} = 0,33M
\end{array}\)
\(\begin{array}{l}
c)\\
{m_{NaCl}} = 5,85g\\
d)\\
C{\% _{NaCl}} = \dfrac{{5,85}}{{288,3}} \times 100\% = 2,03\%
\end{array}\)