Đáp án:

Câu 11: 

\(\begin{array}{l}
C{\% _{NaOH}} = 19,2\% \\
C{\% _{N{a_2}C{O_3}}} = 16,86\% \\
C{\% _{NaHC{O_3}}} = 8,9\% 
\end{array}\)

Câu 12: 

\(\begin{array}{l}
a,\\
C{\% _{A{l_2}{{(S{O_4})}_3}}} = 24,76\% \\
C{\% _{{H_2}S{O_4}(dư)}} = 2,36\% \\
b,\\
C{\% _{NaOH(dư)}} = 0,92\% \\
C{\% _{N{a_2}S{O_4}}} = 16,4\% 
\end{array}\)

Giải thích các bước giải:

Câu 11: 

\(\begin{array}{l}
{n_{C{O_2}}} = 0,6mol\\
C{O_2} + 2NaOH \to N{a_2}C{O_3} + {H_2}O(1)\\
C{O_2} + NaOH \to NaHC{O_3}(2)
\end{array}\)

Gọi a và b lần lượt là số mol của CO2(1) và CO2(2)

\(\begin{array}{l}
\left\{ \begin{array}{l}
a + b = 0,6\\
106a + 84b = 58,3
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,36\\
b = 0,24
\end{array} \right.\\
 \to {n_{C{O_2}(1)}} = 0,36mol\\
 \to {n_{C{O_2}(2)}} = 0,24mol\\
 \to {n_{NaOH}} = 2{n_{C{O_2}(1)}} + {n_{C{O_2}(2)}} = 0,96mol\\
 \to {m_{NaOH}} = 38,4g\\
 \to C{\% _{NaOH}} = \dfrac{{38,4}}{{200}} \times 100\%  = 19,2\% \\
{n_{N{a_2}C{O_3}}} = {n_{C{O_2}(1)}} = 0,36mol\\
 \to {m_{N{a_2}C{O_3}}} = 38,16g\\
{n_{NaHC{O_3}}} = {n_{C{O_2}(2)}} = 0,24mol\\
 \to {m_{NaHC{O_3}}} = 20,16g\\
{m_{{\rm{dd}}A}} = {m_{C{O_2}}} + {m_{{\rm{dd}}NaOH}} = 226,4g\\
 \to C{\% _{N{a_2}C{O_3}}} = \dfrac{{38,16}}{{226,4}} \times 100\%  = 16,86\% \\
 \to C{\% _{NaHC{O_3}}} = \dfrac{{20,16}}{{226,4}} \times 100\%  = 8,9\% 
\end{array}\)

Câu 12:

\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
A{l_2}{(S{O_4})_3} + 6NaOH \to 2Al{(OH)_3} + 3N{a_2}S{O_4}\\
{H_2}S{O_4} + 2NaOH \to N{a_2}S{O_4} + 2{H_2}O\\
a,\\
{m_{{H_2}S{O_4}}} = \dfrac{{200 \times 24,5\% }}{{100\% }} = 49g\\
 \to {n_{{H_2}S{O_4}}} = 0,5mol\\
{n_{{H_2}}} = 0,45mol\\
 \to \dfrac{{{n_{{H_2}S{O_4}}}}}{3} > \dfrac{{{n_{{H_2}}}}}{3} \to {n_{{H_2}S{O_4}}}dư\\
 \to {n_{{H_2}S{O_4}(pt)}} = {n_{{H_2}}} = 0,45mol\\
 \to {n_{{H_2}S{O_4}(dư)}} = 0,05mol \to {m_{{H_2}S{O_4}(dư)}} = 4,9g\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,3mol \to {m_{Al}} = 8,1g\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{1}{3}{n_{{H_2}}} = 0,15mol\\
 \to {m_{A{l_2}{{(S{O_4})}_3}}} = 51,3g\\
{m_{{\rm{dd}}A}} = {m_{{\rm{dd}}{H_2}S{O_4}}} + {m_{Al}} - {m_{{H_2}}} = 207,2g\\
 \to C{\% _{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{51,3}}{{207,2}} \times 100\%  = 24,76\% \\
 \to C{\% _{{H_2}S{O_4}(du)}} = \dfrac{{4,9}}{{207,2}} \times 100\%  = 2,36\% \\
b,\\
{m_{NaOH}} = \dfrac{{400 \times 11\% }}{{100\% }} = 44g\\
 \to {n_{NaOH}} = 1,1mol\\
{n_{NaOH(pt)}} = 6{n_{A{l_2}{{(S{O_4})}_3}}} + 2{n_{{H_2}S{O_4}(du)}} = 1mol\\
 \to {n_{NaOH(dư)}} = 0,1mol\\
 \to {m_{NaOH(dư)}} = 4g\\
{n_{N{a_2}S{O_4}}} = 3{n_{A{l_2}{{(S{O_4})}_3}}} + {n_{{H_2}S{O_4}(dư)}} = 0,5mol\\
 \to {m_{N{a_2}S{O_4}}} = 71g\\
{n_{Al{{(OH)}_3}}} = 2{n_{A{l_2}{{(S{O_4})}_3}}} = 0,3mol\\
 \to {m_{Al{{(OH)}_3}}} = 23,4g\\
{m_{{\rm{dd}}C}} = {m_{A{l_2}{{(S{O_4})}_3}}} + {m_{{H_2}S{O_4}(dư)}} + {m_{{\rm{dd}}NaOH}} - {m_{Al{{(OH)}_3}}} = 432,8g\\
 \to C{\% _{NaOH(dư)}} = \dfrac{4}{{432,8}} \times 100\%  = 0,92\% \\
 \to C{\% _{N{a_2}S{O_4}}} = \dfrac{{71}}{{432,8}} \times 100\%  = 16,4\% 
\end{array}\)