$n_{CO2}$ = $\frac{22,4}{22,4}$ = `1(mol)`

a) `PTHH:` $C_{6}$$H_{12}$$O_{6}$ `->`$^{Men rượu/30-35°C}$ `2`$C_{2}$$H_{5}$$OH_{}$ + `2`$CO_{2}$ 

b)  $C_{6}$$H_{12}$$O_{6}$ `->`$^{Men rượu/30-35°C}$ `2`$C_{2}$$H_{5}$$OH_{}$ + `2`$CO_{2}$ 

`PỨ` :   `0,5`           `<-`                  `1`             `<-`       `1`        (mol)

$m_{C2H5OH}$ = `1.46 = 46 (g)`

c) $mlí thuyết_{C6H12O6}$ = `0,5.180 = 90(g)`

d) $mthực tế_{C6H12O6}$ = $\frac{90.100}{70}$ = `128,57(g)` 

Đáp án + Giải thích các bước giải:

 Câu 11:
a) PTHH:
$C_6H_{12}O_6 \xrightarrow{\text{men rượu} 30^oC - 35^oC} 2C_2H_5OH + 2CO_2 \uparrow$
b) $n_{CO_2} =\dfrac{V}{22,4} = \dfrac{22,4}{22,4} = 1\text{ mol}$

Theo PTHH:
$n_{C_2H_5OH} = n_{CO_2} = 1\text{ mol}$
$\Rightarrow m_{C_2H_5OH} =n .M = 1 . 46 = 46g$
c) Theo PTHH:
$n_{C_6H_{12}O_6} = \dfrac{1}{2}n_{CO_2} = \dfrac{1}{2} . 1 = 0,5\text{ mol}$

$\Rightarrow m_{C_6H_{12}O_6} = n . M = 0,5 . 180 = 90g$

d) $n_{C_6H_{12}O_6(tt)} = 0,5 : 70\% = \dfrac{5}{7}\text{ mol}$
$\Rightarrow m_{C_6H_{12}O_6(tt)}  = n . M = \dfrac{5}{7} . 180 \approx 128,57g$