Đáp án:
\(\begin{array}{l}
a){V_{S{O_2}}} = 0,0224l\\
b)C{M_{N{a_2}S{O_4}}} = 0,002M\\
c)C{\% _{N{a_2}S{O_4}}} = 11,04\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
N{a_2}S{O_3} + {H_2}S{O_4} \to N{a_2}S{O_4} + S{O_2} + {H_2}O\\
a)\\
{m_{Na{a_2}S{O_3}}} = \dfrac{{0,35 \times 36\% }}{{100\% }} = 0,126g\\
\to {n_{Na{a_2}S{O_3}}} = 0,001mol\\
\to {n_{S{O_2}}} = {n_{Na{a_2}S{O_3}}} = 0,001mol\\
\to {V_{S{O_2}}} = 0,0224l\\
b)\\
{n_{N{a_2}S{O_4}}} = {n_{Na{a_2}S{O_3}}} = 0,001mol\\
\to C{M_{N{a_2}S{O_4}}} = \dfrac{{0,001}}{{0,5}} = 0,002M\\
c)\\
{n_{{H_2}S{O_4}}} = {n_{Na{a_2}S{O_3}}} = 0,001mol\\
\to {m_{{H_2}S{O_4}}} = 0,098g\\
\to {m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,098}}{{9,8\% }} \times 100\% = 1g\\
{m_{N{a_2}S{O_4}}} = 0,142g\\
{m_{{\rm{dd}}}} = {m_{{\rm{dd}}N{a_2}S{O_3}}} + {m_{{\rm{dd}}{H_2}S{O_4}}} - {m_{S{O_2}}} = 1,286g\\
\to C{\% _{N{a_2}S{O_4}}} = \dfrac{{0,142}}{{1,286}} \times 100\% = 11,04\%
\end{array}\)
