Question

5?

Answer 1

Câu 5:

$a\big)$

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

$\to 94x+46y=14(1)$

$n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)$

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

$C_2H_5OH+Na\to C_6H_5ONa+\dfrac{1}{2}H_2$

Theo PT: $x+y=0,2(2)$

Từ $(1)(2)\to x=y=0,1(mol)$

$\to \%m_{C_6H_5OH}=\dfrac{0,1.94}{14}.100\%\approx 67,14\%$

$\to \%m_{C_2H_5OH}=100-67,14=32,86\%$

$b\big)$

$C_6H_5OH+3HNO_3\to C_6H_2(NO_2)_3OH+3H_2O$

Theo PT: $n_{C_6H_2(NO_2)_3OH}=x=0,1(mol)$

$\to m_{C_6H_2(NO_2)_3OH}=0,1.229=22,9(g)$