Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{N{a_2}S{O_3}}} = 64\% \\
\% {m_{N{a_2}S{O_4}}} = 36\% \\
b)\\
C{\% _{HCl}} = 2,43\% \\
C{M_{HCl}} = 0,8M\\
c)\\
C{\% _{NaCl}} = 3,73\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
N{a_2}S{O_3} + 2HCl \to 2NaCl + S{O_2} + {H_2}O\\
a)\\
{n_{S{O_2}}} = 0,2mol\\
\to {n_{N{a_2}S{O_3}}} = {n_{S{O_2}}} = 0,2mol\\
\to {m_{N{a_2}S{O_3}}} = 25,2g\\
\to {m_{N{a_2}S{O_4}}} = 14,2g\\
\to \% {m_{N{a_2}S{O_3}}} = \dfrac{{25,2}}{{39,4}} \times 100\% = 64\% \\
\to \% {m_{N{a_2}S{O_4}}} = 100\% - 64\% = 36\% \\
b)\\
{n_{HCl}} = 2{n_{S{O_2}}} = 0,4mol\\
\to {m_{HCl}} = 14,6g\\
\to C{\% _{HCl}} = \dfrac{{14,6}}{{600}} \times 100\% = 2,43\% \\
{V_{HCl}} = \dfrac{{600}}{{1,2}} = 500ml = 0,5l\\
\to C{M_{HCl}} = \dfrac{{0,4}}{{0,5}} = 0,8M\\
c)\\
{n_{NaCl}} = 2{n_{S{O_2}}} = 0,4mol\\
{m_{NaCl}} = 23,4g\\
{m_{{\rm{dd}}}} = {m_{hh}} + {m_{{\rm{dd}}HCl}} - {m_{S{O_2}}} = 626,6g\\
\to C{\% _{NaCl}} = \dfrac{{23,4}}{{626,6}} \times 100\% = 3,73\%
\end{array}\)
