Đáp án:

 a) 14,1 g và 6,9g

b) 150 ml

c) 34,35 g

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
2{C_6}{H_5}OH + 2Na \to 2{C_6}{H_5}ONa + {H_2}\\
n{H_2} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol \Rightarrow nhh = 0,15 \times 2 = 0,3\,mol\\
hh:{C_2}{H_5}OH(a\,mol),{C_6}{H_5}OH(b\,mol)\\
46a + 94b = 21\\
a + b = 0,3\\
 \Rightarrow a = b = 0,15\,mol\\
m{C_2}{H_5}OH = 0,15 \times 46 = 6,9g\\
m{C_6}{H_5}OH = 21 - 6,9 = 14,1g\\
b)\\
{C_6}{H_5}OH + NaOH \to {C_6}{H_5}ONa + {H_2}O\\
nNaOH = n{C_6}{H_5}OH = 0,15\,mol\\
VNaOH = \dfrac{{0,15}}{1} = 0,15l = 150\,ml\\
c)\\
{C_6}{H_5}OH + 3HN{O_3} \to {C_6}{H_2}OH{(N{O_2})_3} + 3{H_2}O\\
n{C_6}{H_2}OH{(N{O_2})_3} = n{C_6}{H_5}OH = 0,15\,mol\\
m{C_6}{H_2}OH{(N{O_2})_3} = 0,15 \times 229 = 34,35g
\end{array}\)