\(a,n_{CH_3COOC_2H_5}=\dfrac{11}{88}=0,125\left(mol\right)\)

PTHH: \(CH_3COOH+C_2H_5OH\xrightarrow[t^o]{H_2SO_{4\left(đ\right)}}CH_3COOC_2H+H_2O\)

                0,125<------------------------------0,125

b, => m_CH3COOH = 0,125.60 = 7,5 (g)

\(c,H=\dfrac{7,5}{12}.100\%=62,5\%\)

a.b.\(n_{CH_3COOC_2H_5}=\dfrac{11}{88}=0,125mol\)

\(CH_3COOH+C_2H_5OH\rightarrow\left(t^o,H_2SO_4\left(đ\right)\right)CH_3COOC_2H_5+H_2O\)

   0,125                                                                    0,125             ( mol )

\(m_{CH_3COOH}=0,125.60=7,5g\)

c.\(n_{CH_3COOH}=\dfrac{12}{60}=0,2mol\)

\(H=\dfrac{0,125}{0,2}.100\%=62,5\%\)

\(n_{CH3COOC2H5}=\dfrac{55}{88}=0,625\left(mol\right)\)

a) Pt : \(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\)

               0,625                                      0,625

b) \(m_{CH3COOH}=0,625.60=37,5\left(g\right)\)

 Chúc bạn học tốt

a, PT: \(CH_3COOH+C_2H_5OH\underrightarrow{_{H_2SO_{4\left(đ\right)}}}CH_3COOC_2H_5+H_2O\)

b, Ta có: \(n_{CH_3COOH}=\dfrac{30}{60}=0,5\left(mol\right)\)

\(n_{C_2H_5OH}=\dfrac{27,6}{46}=0,6\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,5}{1}< \dfrac{0,6}{1}\), ta được C₂H₅OH.

Theo PT: \(n_{CH_3COOC_2H_5\left(LT\right)}=n_{CH_3COOH}=0,5\left(mol\right)\)

\(\Rightarrow m_{CH_3COOC_2H_5\left(LT\right)}=0,5.88=44\left(g\right)\)

Mà: m _{CH3COOC2H5 (TT)} = 35,2 (g)

\(\Rightarrow H\%=\dfrac{35,2}{44}.100\%=80\%\)

Bạn tham khảo nhé!

nC2H5OH = 8.05/46 = 0.175 (mol) 

nCH3COOH = 36/60 = 0.6 (mol) 

nCH3COOC2H5 = 12.32/88 = 0.14 (mol) 

C2H5OH + CH3COOH <-H2SO4đ,t⁰-> CH3COOC2H5 + H2O 

1.......................1

0.175................0.6

LTL : 0.175/1 < 0.6/1 

=> CH3COOH dư 

mCH3COOH (dư) = ( 0.6 - 0.175) * 60 = 25.5 (g) 

nCH3COOC2H5 = nC2H5OH = 0.175 (mol) 

H% = 0.14/0.175 * 100% = 80%

Đáp án A

\(n_{CH_3COOC_2H_5}=\dfrac{4,4}{88}=0,05\left(mol\right)\)

PTHH: CH₃COOH + C₂H₅OH --H₂SO₄(đ),t^o--> CH₃COOC₂H₅ + H₂O

               0,05<--------------------------------------0,05

=> \(m_{CH_3COOH\left(lý.thuyết\right)}=0,05.60=3\left(g\right)\)

=> \(m_{CH_3COOH\left(tt\right)}=\dfrac{3.100}{60}=5\left(g\right)\)

\(CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+CO_2+H_2O\\ a,n_{CH_3COOH}=0,2.1=0,2\left(mol\right)\Rightarrow n_{CO_2}=\dfrac{0,2}{2}=0,1\left(mol\right)\Rightarrow V_{CO_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ b,CH_3COOH+C_2H_5OH⇌\left(H^+,t^o\right)CH_3COOC_2H_5+H_2O\\ n_{CH_3COOH}=\dfrac{50}{200}.0,2=0,05\left(mol\right)\\ n_{C_2H_5OH}=\dfrac{23}{46}=0,5\left(mol\right)\\ Vì:\dfrac{0,5}{1}>\dfrac{0,05}{1}\Rightarrow Rượu.dư\\ \Rightarrow n_{este\left(LT\right)}=n_{axit}=0,05\left(mol\right)\\ \Rightarrow n_{este\left(TT\right)}=80\%.0,05=0,04\left(mol\right)\\ m_{CH_3COOC_2H_5}=88.0,04=3,52\left(g\right)\)