Đáp án:

a,

$$n_{O_2}=\dfrac{15,2}{32}=0,475$$ mol

Đốt cháy axit no, đơn chức, mạch hở thu được:

$$n_{H_2O}=n_{CO_2}=\dfrac{10,08}{22,4}=0,45$$ mol 

Bảo toàn khối lượng:

$$m=m_{CO_2}+m_{H_2O}-m_{O_2}=0,45.44+0,45.18-0,475.32=12,7g$$ 

b,

Bảo toàn $$O$$:

$$2n_{\text{axit}}+2n_{CO_2}=2n_{CO_2}+n_{H_2O}$$ 

$$\Rightarrow n_{\text{axit}}=0,2$$ mol 

Đặt $$x, y$$ là số mol $$CH_3COOH, C_2H_5COOH$$ ban đầu 

$$\Rightarrow \begin{cases} x+y=0,2\\ 60x+74y=12,7\end{cases}$$ 

$$\Rightarrow \begin{cases} x= 0,15 \\ y=0,05\end{cases}$$

$$n_{C_2H_5OH}=\dfrac{9,2}{46}=0,2$$ mol 

$$n_{\text{axit}}=n_{C_2H_5OH}$$ nên theo lí thuyết, phản ứng xảy ra vừa đủ  

Đặt $$a, b$$ là số mol $$CH_3COOH, C_2H_5COOH$$ phản ứng 

$$\Rightarrow n_{CH_3COOC_2H_5}=a$$ mol, $$n_{C_2H_5COOC_2H_5}=b$$ mol 

$$\Rightarrow 88a+102b=14,13$$ 

$$n_{\text{axit dư}}=0,2-a-b$$ mol 

$$n_{KOH}=0,045$$ mol 

$$\Rightarrow 0,2-a-b=0,045$$ 

$$\Rightarrow \begin{cases} a=0,12\\ b=0,035\end{cases}$$ 

Vậy:

$$H_{CH_3COOH}=\dfrac{0,12.100\%}{0,15}=80\%$$ 

$$H_{C_2H_5COOH}=\dfrac{0,035.100\%}{0,05}=70\%$$

a)

Đặt n_{CH_3COOH}=x(mol);n_{C_2H_5COOH}=y(mol)

CH_3COOH+2O_2  $$\xrightarrow{t^o}$$  2CO_2+2H_2O

2C_2H_5COOH+7O_2  $$\xrightarrow{t^o}$$  6CO_2+6H_2O

Theo PT:

n_{O_2}=2x+3,5y={15,2}/{32}=0,475(1)

n_{CO_2}=2x+3y={10,08}/{22,4}=0,45(2)

(1)(2)->x=0,15;y=0,05

->m=0,15.60+0,05.74=12,7(g)

b)

n_{C_2H_5OH}={9,2}/{46}=0,2(mol)

Do 0,2=0,15+0,05-> H tính theo cả 3 chất.

CH_3COOH+C_2H_5OH  $$\buildrel{{t^o,H^+}}\over\rightleftharpoons$$  CH_3COOC_2H_5+H_2O

C_2H_5COOH+C_2H_5OH  $$\buildrel{{t^o,H^+}}\over\rightleftharpoons$$  C_2H_5COOC_2H_5+H_2O

CH_3COOH+KOH->CH_3COOK+H_2O

C_2H_5COOH+KOH->C_2H_5COOK+H_2O

45mL=0,045L

Đặt n_{CH_3COOH\ du}=a(mol);n_{C_2H_5COOH\ du}=b(mol)

->n_{CH_3COOH\ pu}=0,15-a(mol);n_{C_2H_5COOH\ pu}=0,05-b(mol)

->n_{CH_3COOC_2H_5}=0,15-a(mol);n_{C_2H_5COOC_2H_5}=0,05-b(mol)

->88(0,15-a)+102(0,05-b)=14,13(2)

n_{KOH}=0,045=a+b(2)

(1)(2)->a=0,03;b=0,015

H_{CH_3COOH}={0,15-0,03}/{0,15}.100\%=80\%

H_{C_2H_5COOH}={0,05-0,015}/{0,05}.100\%=70\%