\(M_{CH_3COOH}=60\left(\dfrac{g}{mol}\right)\)

\(\%C=\dfrac{24}{60}\cdot100\%=40\%\)

\(\%H=\dfrac{4}{60}\cdot100\%=6.67\%\)

\(\%O=100-40-6.67=53.33\%\)

\(a.M_{C_2H_6O}=12,2+2+16=46\left(đvC\right)\\ \%C=\dfrac{12.2}{46}.100=52,17\%\\ \%H=\dfrac{6}{46}.100=13,04\%\\ \%O=100-52,17-13,04=34,79\%\\ b.n_{CO_2}=\dfrac{6.6}{44}=0,15\left(mol\right)\\ BTNT\left(C\right):n_{C_2H_6O}.2=n_{CO_2}.1\\ \Rightarrow n_{C_2H_6O}=0,075\left(mol\right)\\ \Rightarrow m_{C_2H_6O}=0,075.46=3,45\left(g\right)\)

\(a,\%m_C=\dfrac{12.2}{12.2+6.1+16}.100\approx52,174\%\\ \%m_H=\dfrac{6.1}{12.2+6.1+16}.100\approx13,043\%\\ \%m_O=\dfrac{16}{12.2+6.1+16}.100\approx34,783\%\)

\(b,n_C=n_{CO_2}=\dfrac{6,6}{44}=0,15\left(mol\right)\\ \Rightarrow m_{C_2H_5OH}=\dfrac{n_C}{2}=\dfrac{0,15}{2}=0,075\left(mol\right)\\ \Rightarrow m_{C_2H_5OH}=0,075.46=3,45\left(g\right)\)

bạn có cần câu trả lời nữa không?

công thức tổng quát:CxHyOz

x= \(\frac{60.40}{12.100}\)= 2

y= \(\frac{60.6,67}{1.100}\)= 4

%O= 100-40-6,67= 53,33%

z= \(\frac{60.53,33}{16.100}\)= 2

=> CTHH: C₂H₄O₂

ông lớp 9 à :))

\(n_{CH_3COOH}=\dfrac{4,8}{60}=0,08\left(mol\right)\)

\(V_{C_2H_5OH}=11,5.45\%=5,175\left(ml\right)\\ \rightarrow m_{C_2H_5OH}=5,175.0,8=4,14\left(g\right)\\ \rightarrow n_{C_2H_5OH}=\dfrac{4,14}{46}=0,09\left(mol\right)\)

\(PTHH:C_2H_5OH+CH_3COOH\underrightarrow{H_2SO_4đ,t^o}CH_3COOC_2H_5+H_2O\)

LTL: 0,09 > 0,08 => C₂H₅OH dư

Đề thiếu à :) ?

\(n_{HCOOH} = a(mol) ; n_{CH_3COOH} =b(mol)\\ \Rightarrow 46a + 60b = 15,2(1)\\ HCOOH + NaHCO_3 \to HCOONa + CO_2 + H_2O\\ CH_3COOH + NaHCO_3 \to CH_3COONa + CO_2 + H_2O\\ n_{CO_2} = a + b = \dfrac{3,36}{22,4} = 0,15(2)\\ (1)(2) \Rightarrow a = \dfrac{-31}{70}<0\)

(Sai đề)

Sửa đề : Na₂CO₃

\(n_{HCOOH}=a\left(mol\right),n_{CH_3COOH}=b\left(mol\right)\)

\(\Rightarrow m_{hh}=46a+60b=15.2\left(g\right)\left(1\right)\)

\(2HCOOH+Na_2CO_3\rightarrow2HCOONa+CO_2+H_2O\)

\(a......................................0.5a\)

\(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)

\(b...........................................0.5b\)

\(n_{CO_2}=0.5a+0.5b=\dfrac{3.36}{22.4}=0.15\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.2,b=0.1\)

\(m_{HCOOH}=0.2\cdot46=9.2\left(g\right)\)

\(m_{CH_3COOH}=6\left(g\right)\)

Chú ý lần sau viết đề cẩn thận hơn nha em ơi !

Bài 1:

PTHH: \(C_2H_5OH+O_2\xrightarrow[]{mengiấm}CH_3COOH+H_2O\)

Ta có: \(n_{C_2H_5OH}=\dfrac{115\cdot0,8}{46}=2\left(mol\right)=n_{CH_3COOH\left(lýthuyết\right)}\)

\(\Rightarrow m_{CH_3COOH\left(thực\right)}=2\cdot60\cdot90\%=108\left(g\right)\)

Bài 2:

PTHH: \(C_2H_5OH+CH_3COOH\xrightarrow[H_2SO_4\left(đ\right)]{t^o}CH_3COOC_2H_5+H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{CH_3COOH}=\dfrac{60}{60}=1\left(mol\right)\\n_{C_2H_5OH}=\dfrac{92}{46}=2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Rượu còn dư, Axit p/ứ hết

\(\Rightarrow n_{CH_3COOC_2H_5\left(lýthuyết\right)}=1\left(mol\right)\) \(\Rightarrow m_{CH_3COOC_2H_5\left(thực\right)}=1\cdot88\cdot80\%=70,4\left(g\right)\)

Axit axetic \(CH_3COOH\)

Rượu etylic \(C_2H_5OH\)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)

\(2CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+H_2\)

0,2                                                               0,1

\(m_{CH_3COOH}=0,2\cdot60=12g\)

\(\%m_{CH_3COOH}=\dfrac{12}{20}\cdot100\%=60\%\)

\(\%m_{C_2H_5OH}=100\%-60\%=40\%\)

Bài 4. Axit axetic có công thức C₂H₄0₂. Hãy tính thành phần phần trăm khối lượng của các nguyên tố trong axit axetic.

Lời giải:

Từ C₂H₄O₂ ta có: M = 60 g/mol; m_C = 2 x 12 = 24 g; m_H = 4 x 1 = 4 g;

M_O = 2 x 16 = 32 g.

%C = (24 : 60) x 100% = 40%; %H = (4 : 60) x 100% = 6,67%;

%O = 100% - 40% - 6,67% = 53,33%.