\(n_{CO2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Pt : \(C_6H_{12}O_6\xrightarrow[30-35^oC]{Menrượu}2C_2H_5OH+2CO_2\)

                                           0,5            0,5

a) \(m_{C2H5OH}=0,5.46=23\left(g\right)\)

b) Pt : \(C_2H_5OH+O_2\xrightarrow[]{Mengiấm}CH_3COOH+H_2O\)

                0,5                                     0,5

\(m_{CH3COOH\left(lt\right)}=0,5.60=30\left(g\right)\)

⇒ \(m_{CH3COOH\left(tt\right)}=30.80\%=24\left(g\right)\)

 Chúc bạn học tốt

\(C_6H_{12}O_6\underrightarrow{t^o}2C_2H_5OH+2CO_2\uparrow\)(xt : men rượu )

                          0,5          0,5

\(n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

\(m_{C_2H_5OH}=0,5.46=23\left(g\right)\)

\(C_2H_5OH+O_2\underrightarrow{t^o}CH_3COOH+H_2O\) (men giấm )

0,5                                 0,5

\(m_{CH_3COOH}=0,5.60=30\left(g\right)\)

\(m_{CH_3COOHtt}=30.80\%=24\left(g\right)\)

a) n glucozo = 54/180 = 0,3(mol)

n glucozo pư = 0,3.80% = 0,24(mol)

$C_6H_{12}O_6 \xrightarrow{t^o} 2CO_2 +2 C_2H_5OH$

n C2H5OH = 2n glucozo = 0,48(mol)

m C2H5OH = 0,48.46 = 22,08(gam)

b)

$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
n CH3COOH = n C2H5OH = 0,48(mol)

C% CH3COOH = 0,48.60/500  .100% = 5,76%

Câu 8:

\(n_{C_2H_5OH}=\dfrac{4,6}{46}=0,1\left(mol\right)\)

PT: \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)

Theo PT: \(n_{CH_3COOH\left(LT\right)}=n_{C_2H_5OH}=0,1\left(mol\right)\)

\(\Rightarrow m_{CH_3COOH\left(LT\right)}=0,1.60=6\left(g\right)\)

\(\Rightarrow H=\dfrac{4,8}{6}.100\%=80\%\)

Đáp án: A

B.40%

\(a,n_{C_6H_{12}O_6}=\dfrac{450}{180}=2,5\left(mol\right)\)

PTHH: C₆H₁₂O₆ --men rượu--> 2C₂H₅OH + 2CO₂

                2,5-------------------------->5

            C₂H₅OH + O₂ --men giấm--> CH₃COOH + H₂O

               5------------------------------------>5

\(b,m_{C_2H_5OH}=5.46=230\left(g\right)\)

\(c,m_{CH_3COOH}=5.80\%.60=240\left(g\right)\)

a)\(n_{C_6H_{12}O_6}=\dfrac{450}{180}=2,5mol\)

\(C_6H_{12}O_6\underrightarrow{menrượu}2C_2H_5OH+2CO_2\)

2,5                              5

b)\(m_{C_2H_5OH}=5\cdot46=230g\)

c)\(C_2H_5OH+O_2\underrightarrow{mengiấm}CH_3COOH+H_2O\)

    5                                         5

Thực tế: \(n_{CH_3COOH}=5\cdot80\%=4mol\)

\(m_{CH_3COOH}=4\cdot60=240g\)

Đáp án: C

m g l u c o z ơ   n g u y ê n   c h ấ t   =   5 . 80 %   =   4   k g   V ì   h i ệ u   s u ấ t   p h ả n   ứ n g   đ ạ t   90 %   = >   m g l u c o z ơ   =   4 . 90 %   =   3 , 6   k g

                        C 6 H 12 O 6   → m e n   r ư ợ u     2 C 2 H 5 O H   +   2 C O 2

P T :               180   k g                                             2 . 46   k g   P ứ :                 3 , 6   k g                   →               3 , 6 . 2 . 46 180 = 1 , 84   k g

= >   m r ư ợ u   e t y l i c   t h u   đ ư ợ c   =   1 , 84   k g   =   1840   g a m

\(m_{rượu} = 10.1000.0,8 = 8000(gam)\\ n_{rượu} = \dfrac{8000}{46} = \dfrac{4000}{23}(mol)\\ C_2H_5OH + O_2 \xrightarrow{t^o} CH_3COOH + H_2O\\ n_{CH_3COOH} = n_{C_2H_5OH} =\dfrac{4000}{23}.80\% = \dfrac{3200}{23}(mol)\\ C_{M_{CH_3COOH}} = \dfrac{\dfrac{3200}{23}}{10} =13,91M \)