Đáp án:
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Giải thích các bước giải:
\(\begin{array}{l}
2.\\
A{l_2}{O_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
{n_{A{l_2}{O_3}}} = 0,2mol\\
a.\\
{n_{{H_2}S{O_4}}} = 3{n_{A{l_2}{O_3}}} = 0,6mol\\
\to C{M_{{H_2}S{O_4}}} = \dfrac{{0,6}}{{0,2}} = 3M\\
b.\\
{n_{A{l_2}{{(S{O_4})}_3}}} = {n_{A{l_2}{O_3}}} = 0,2mol\\
\to C{M_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{0,2}}{{0,2}} = 1M\\
3.\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{{H_2}}} = 0,3mol\\
a.\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,2mol\\
\to {m_{Al}} = 5,4g\\
b.\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,6mol\\
\to {m_{HCl}} = 21,9g\\
\to {m_{{\rm{dd}}HCl}} = \dfrac{{21,9}}{{7,3\% }} \times 100\% = 300g\\
c.\\
{n_{AlC{l_3}}} = \dfrac{2}{3}{n_{{H_2}}} = 0,2mol\\
\to {m_{AlC{l_3}}} = 26,7g\\
\to {m_{{\rm{dd}}}} = {m_{Al}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 304,8g\\
\to C{\% _{AlC{l_3}}} = \dfrac{{26,7}}{{304,8}} \times 100\% = 8,8\% \\
4.\\
BaC{l_2} + N{a_2}C{O_3} \to 2NaCl + BaC{O_3}\\
{m_{BaC{l_2}}} = \dfrac{{100 \times 20,8\% }}{{100\% }} = 20,8g\\
\to {n_{BaC{l_2}}} = 0,1mol\\
a.\\
{n_{BaC{O_3}}} = {n_{BaC{l_2}}} = 0,1mol\\
\to {m_{BaC{O_3}}} = 19,7g\\
b.\\
{n_{N{a_2}C{O_3}}} = {n_{BaC{l_2}}} = 0,1mol\\
\to {m_{N{a_2}C{O_3}}} = 10,6g\\
\to C{\% _{N{a_2}C{O_3}}} = \dfrac{{10,6}}{{200}} \times 100\% = 5,3\% \\
c.\\
{n_{NaCl}} = 2{n_{BaC{l_2}}} = 0,2mol\\
\to {m_{NaCl}} = 11,7g\\
{m_{{\rm{dd}}}} = {m_{{\rm{dd}}BaC{l_2}}} + {m_{{\rm{dd}}N{a_2}C{O_3}}} - {m_{BaC{O_3}}} = 280,3g\\
\to C{\% _{NaCl}} = \dfrac{{11,7}}{{280,3}} \times 100\% = 4,2\% \\
5.\\
{\rm{CuS}}{O_4} + 2NaOH \to N{a_2}S{O_4} + Cu{(OH)_2}\\
{m_{{\rm{CuS}}{O_4}}} = \dfrac{{150 \times 16\% }}{{100\% }} = 24g\\
\to {n_{{\rm{CuS}}{O_4}}} = 0,15mol\\
a.\\
{n_{Cu{{(OH)}_2}}} = {n_{{\rm{CuS}}{O_4}}} = 0,15mol\\
\to {m_{Cu{{(OH)}_2}}} = 14,7g\\
b.\\
{n_{NaOH}} = 2{n_{{\rm{CuS}}{O_4}}} = 0,3mol\\
\to {m_{NaOH}} = 12g\\
\to C{\% _{NaOH}} = \dfrac{{12}}{{200}} \times 100\% = 6\% \\
c.\\
{n_{N{a_2}S{O_4}}} = {n_{{\rm{CuS}}{O_4}}} = 0,15mol\\
\to {m_{N{a_2}S{O_4}}} = 21,3g\\
{m_{{\rm{dd}}}} = {m_{{\rm{ddCuS}}{{\rm{O}}_4}}} + {m_{{\rm{dd}}NaOH}} - {m_{Cu{{(OH)}_2}}} = 335,3g\\
\to C{\% _{N{a_2}S{O_4}}} = \dfrac{{21,3}}{{335,3}} \times 100\% = 6,4\%
\end{array}\)
