Đáp án + Giải thích các bước giải:

$a\big)$

Đặt $n_{C_6H_5OH}=x(mol);n_{C_2H_5OH}=y(mol)$

$\to 94x+46y=32,6(1)$

$C_6H_5OH+Na\to C_6H_5ONa+\frac{1}{2}H_2$

$C_2H_5OH+Na\to C_2H_5ONa+\frac{1}{2}H_2$

Theo PT: $x+y=2n_{H_2}=2.\frac{5,6}{22,4}=0,5(2)$

$(1)(2)\to x=0,2;y=0,3$

$\to m_{C_6H_5OH}=0,2.94=18,8(g);m_{C_2H_5OH}=0,3.46=13,8(g)$

$b\big)$

$C_6H_5OH+3HNO_3\xrightarrow{H_2SO_{4(đ)}}C_6H_2(NO_2)_3OH↓+3H_2O$

Theo PT: $n_{C_6H_2(NO_2)_3OH}=n_{C_6H_5OH}=0,2(mol)$

$\to m=m_{C_6H_2(NO_2)_3OH}=0,2.229=45,8(g)$

Em tham khảo nha :

\(\begin{array}{l}
a)\\
2{C_6}{H_5}OH + 2Na \to 2{C_6}{H_5}ONa + {H_2}\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
{n_{{H_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25mol\\
hh:{C_6}{H_5}OH(a\,mol),{C_2}{H_5}OH(b\,mol)\\
\left\{ \begin{array}{l}
94a + 46b = 32,6\\
\dfrac{1}{2}a + \dfrac{1}{2}b = 0,25
\end{array} \right.\\
 \Rightarrow a = 0,2;b = 0,3\\
{m_{{C_6}{H_5}OH}} = 0,2 \times 94 = 18,8g\\
{m_{{C_2}{H_5}OH}} = 32,6 - 18,8 = 13,8g\\
b)\\
{C_6}{H_5}OH + 3HN{O_3} \to {C_6}{H_2}{(N{O_2})_3}OH + 3{H_2}O\\
{n_{{C_6}{H_2}{{(N{O_2})}_3}OH}} = {n_{{C_6}{H_5}OH}} = 0,2mol\\
{m_{cr}} = 0,2 \times 229 = 45,8g
\end{array}\)