Hệ nghiệm âm.

Đáp án:

Giải thích các bước giải:

Gọi $n_{\text{etylen glicol}} = a(mol) ; n_{\text{ancol etylic}} = b(mol) ; n_{glixerol} = c(mol)$
$C_2H_4(OH)_2 + \frac{5}{2}O_2 \xrightarrow{t^o} 2CO_2 + 3H_2O$
$C_2H_5OH + 3O_2 \xrightarrow{t^o} 2CO_2 + 3H_2O$
$C_3H_5(OH)_3 + \frac{7}{2}O_2 \xrightarrow{t^o} 3CO_2 + 4H_2O$
Ta có :
$n_{CO_2} = 2n_{C_2H_4(OH)_2} + 2n_{C_2H_5OH} + 3n_{C_3H_5(OH)_3}$
$⇒ 2a + 2b + 3c = \frac{15,68}{22,4}= 0,7(1)$
$n_{H_2O} = 3n_{C_2H_4(OH)_2} + 3n_{C_2H_5OH} + 4n_{C_3H_5(OH)_3}$
$⇒ 3a + 3b + 4c = \frac{18}{18} = 1(2)$
có : $n_{Cu(OH)_2} = \frac{29,4}{98}= 0,3(mol)$
$2C_2H_4(OH)_2 + Cu(OH)_2 → [C_2H_4(OH)O]_2Cu + 2H_2O$
$2C_3H_5(OH)_3 + Cu(OH)_2 → [C_3H_5(OH)_2O]_2Cu + 2H_2O$
$⇒ n_{Cu(OH)_2} = 0,5a + 0,5c(mol)$
$⇒ \dfrac{0,5a+0,5c}{62a+46b+92c} = \dfrac{0,3}{80}(3)$
Từ (1) (2) và (3) suy ra $a = 0,043 ; b = 0,157 ; c= 0,1$
$⇒ m = 0,043.62 + 0,157.46 + 0,1.92 = 19,088(gam)$
Vậy :
$\%m_{C_2H_4(OH)_2} = \frac{0,043.62}{19,088}.100\% = 14\%$
$\%m_{C_2H_5OH} = \frac{0,157.46}{19,088}.100\% = 37,84\%$
$⇒ \%m_{C_3H_5(OH)_3} = 100\% - 14\% - 37,84\% = 51,16\%$