Câu `9` 

$n_{C_2H_6O}=\dfrac{46}{46}=1(mol)$ 

$\rm PTHH: C_2H6O +3O_2 \xrightarrow{to} 2CO_2 +3H_2O$ 

theo pthh 

$n_{CO_2}=2n_{C_2H_6O}=2(mol)$ 

`->`$V_{CO_2}=2.22,4=44,8(l)$

$n_{H_2O}=3n_{C_2H_6O}=3(mol)$ 

`->`$m_{H_2O}=3.18 =54(g)$ 

Câu `10`

$(\kern-6pt- C_6H_{10}O_5-\kern-6pt)_n +nH_2O \xrightarrow{acid,to} nC_6H_12O_6 $

$\rm PTHH: C_6H_{12}O_6 \xrightarrow{men-rượu } 2C_2H_5OH +2CO_2$ 

Câu 9:

$\rm n_{C_2H_6O}=\dfrac{46}{46}=1\ (mol)$

Phản ứng xảy ra:

$\rm C_2H_6O+3O_2\xrightarrow{t^°}2CO_2+3H_2O$

Theo phương trình:

$\rm n_{CO_2}=2n_{C_2H_6O}=2\ (mol)\\n_{H_2O}=3n_{C_2H_6O}=3\ (mol)$

$\rm V_{CO_2}=2×22,4=44,8\ (L)\\m_{H_2O}=3×18=54\ (g)$

Câu 10:

Phương trình điều chế:

$\rm (\kern-6pt- C_6H_{10}O_5 -\kern-6pt)_n+nH_2O\xrightarrow{Acid,\ t^°}nC_6H_{12}O_6\\C_6H_{12}O_6\xrightarrow{enzyme ,\ t^°}2C_2H_5OH+2CO_2$