Đáp án:
a) 36,64 g
b) 22,9g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
2{C_6}{H_5}OH + 2Na \to 2{C_6}{H_5}ONa + {H_2}\\
n{H_2} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol \Rightarrow nhh = 0,2 \times 2 = 0,4\,mol\\
{C_6}{H_5}OH + KOH \to {C_6}{H_5}OK + {H_2}O\\
nKOH = \dfrac{{21,28}}{{56}} = 0,38\,mol\\
n{C_6}{H_5}OH = nKOH = 0,38\,mol\\
n{C_2}{H_5}OH = 0,4 - 0,38 = 0,02\,mol\\
mX = 0,38 \times 94 + 0,02 \times 46 = 36,64g\\
b)\\
{C_6}{H_5}OH + 3HN{O_3} \to {C_6}{H_2}OH{(N{O_2})_3} + 3{H_2}O\\
n{C_6}{H_2}OH{(N{O_2})_3} = n{C_6}{H_5}OH = 0,1\,mol\\
m{C_6}{H_2}OH{(N{O_2})_3} = 0,1 \times 229 = 22,9g
\end{array}\)