6/
a, $C_2H_6O+ 3O_2 \buildrel{{t^o}}\over\to 2CO_2+ 3H_2O$
b,
$n_{CO_2}= \frac{5,6}{22,4}= 0,25 mol$
=> $n_{C_2H_6O}= 0,125 mol$
$H= 90\%$ nên cần $n_{C_2H_5OH}= 0,125:90\%= \frac{5}{36}$
=> $m_{C_2H_5OH}= 46.\frac{5}{36}= 6,4g$
c, $C_2H_4+ H_2O \buildrel{{HCl}}\over\to C_2H_5OH$
=> $n_{C_2H_4}= \frac{5}{36}$
=> $V_{C_2H_4}= 22,4.\frac{5}{36}= 3,1l$
7/
a,
$n_{C_2H_4}= 1 mol$
$C_2H_4+ H_2O \buildrel{{HCl}}\over\to C_2H_5OH$
Theo lí thuyết tạo 1 mol rượu
$n_{C_2H_5OH}= \frac{13,8}{46}= 0,3 mol$
=> $H= \frac{0,3.100}{1}= 30\%$
b,
$C_2H_6O+ 3O_2 \to 2CO_2+ 3H_2O$
=> $n_{O_2}= 0,9 mol$
=> $V_{O_2}= 0,9.22,4= 20,16l$
