Đáp án:

Bạn tham khảo nha ! 

Giải thích các bước giải:

 \(\begin{array}{l}
1.\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{m_{NaOH}} = 1g\\
 \to {n_{NaOH}} = 0,025mol\\
{V_{{H_2}S{O_4}}} = \dfrac{{51}}{{1,02}} = 50ml\\
 \to {n_{{H_2}S{O_4}}} = 0,1mol\\
 \to {n_{NaOH}} < {n_{{H_2}S{O_4}}} \to {n_{{H_2}S{O_4}}}dư
\end{array}\)

\(\begin{array}{l}
a,\\
{n_{N{a_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = \dfrac{1}{2}{n_{NaOH}} = 0,0125mol\\
 \to {n_{{H_2}S{O_4}(dư)}} = 0,0875mol\\
 \to {m_{{H_2}S{O_4}(dư)}} = 8,575g\\
 \to {m_{N{a_2}S{O_4}}} = 1,775g\\
{m_{{\rm{dd}}}} = {m_{{\rm{dd}}NaOH}} + {m_{{\rm{dd}}{H_2}S{O_4}}} = 76g\\
 \to C{\% _{{H_2}S{O_4}(dư)}} = \dfrac{{8,575}}{{76}} \times 100\%  = 11,28\% \\
 \to C{\% _{N{a_2}S{O_4}}} = \dfrac{{1,775}}{{76}} \times 100\%  = 2,34\% 
\end{array}\)

b, Bạn xem lại đề nha.

\(\begin{array}{l}
2.\\
Na + HCl \to NaCl + \frac{1}{2}{H_2}\\
{n_{Na}} = 0,5mol\\
{m_{HCl}} = \dfrac{{100 \times 7,3\% }}{{100\% }} = 7,3g\\
 \to {n_{HCl}} = 0,2mol\\
 \to {n_{Na}} > {n_{HCl}} \to {n_{Na}}dư
\end{array}\)

\(\begin{array}{l}
a,\\
{n_{{H_2}}} = \dfrac{1}{2}{n_{HCl}} = 0,1mol\\
 \to {V_{{H_2}}} = 2,24l\\
b,\\
{n_{NaCl}} = {n_{HCl}} = 0,2mol\\
 \to {m_{NaCl}} = 11,7g\\
{m_{{\rm{dd}}}} = {m_{Na}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 111,3g\\
 \to C{\% _{NaCl}} = \dfrac{{11,7}}{{111,3}} \times 100\%  = 10,51\% 
\end{array}\)

\(\begin{array}{l}
c,\\
AgN{O_3} + NaCl \to NaN{O_3} + AgCl\\
{n_{AgN{O_3}}} = {n_{AgCl}} = {n_{NaCl}} = 0,2mol\\
 \to {V_{AgN{O_3}}} = \dfrac{{0,2}}{1} = 0,2l\\
 \to {m_{AgCl}} = 28,7g
\end{array}\)