Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{{C_3}{H_5}{{(OH)}_3}}} = 70,55\% \\
\% {m_B} = 29,45\% \\
b)\\
CTPT:C{H_4}O
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{P_1}:\\
2{C_n}{H_{2n + 1}}OH + 2Na \to 2{C_n}{H_{2n + 1}}ONa + {H_2}(1)\\
2{C_3}{H_5}{(OH)_3} + 6Na \to 2{C_3}{H_5}{(ONa)_3} + 3{H_2}(2)\\
{P_2}:\\
2{C_3}{H_5}{(OH)_3} + Cu{(OH)_2} \to {{\rm{[}}{C_3}{H_5}{(OH)_2}O]_2}Cu + 2{H_2}O\\
{n_{Cu{{(OH)}_2}}} = \dfrac{{2,45}}{{98}} = 0,025\,mol\\
{n_{{C_3}{H_5}{{(OH)}_3}}} = 2{n_{Cu{{(OH)}_2}}} = 0,05\,mol\\
\% {m_{{C_3}{H_5}{{(OH)}_3}}} = \dfrac{{0,05 \times 2 \times 92}}{{13,04}} \times 100\% = 70,55\% \\
\% {m_B} = 100 - 70,55 = 29,45\% \\
b)\\
{n_{{H_2}}} = \dfrac{{2,352}}{{22,4}} = 0,105\,mol\\
{n_{{H_2}(2)}} = \dfrac{{0,05 \times 3}}{2} = 0,075\,mol\\
{n_{{H_2}(1)}} = 0,105 - 0,075 = 0,03\,mol\\
{n_B} = 2{n_{{H_2}(1)}} = 0,06\,mol\\
{m_B} = 13,04 - 0,1 \times 92 = 3,84g\\
{M_B} = \dfrac{{3,84}}{{0,12}} = 32g/mol \Rightarrow 14n + 18 = 32\\
\Rightarrow n = 1 \Rightarrow CTPT:C{H_4}O
\end{array}\)
