Cho 6g ch3cooh tác dụng 6gc2h5oh ( có axit) sau thu dc ch3cooc2h5 tìm m biết h= 80%

`n_{CH_3COOH}=6/{60}=0,1(mol)`

`n_{C_2H_5OH}=6/{46}\approx 0,13(mol)`

`CH_3COOH+C_2H_5OH`  $\xrightarrow{t^o,H^+}$  `CH_3COOC_2H_5+H_2O`

Do `0,1<0,13->H` tính theo `CH_3COOH.`

`->n_{CH_3COOH\ pu}=0,1.80\%=0,08(mol)`

Theo PT: `n_{CH_3COOC_2H_5\ t t}=n_{CH_3COOH\ pu}=0,08(mol)`

`->m=0,08.88=7,04(g)`

Đáp án:

$\rm n_{CH_3COOH}=\dfrac{6}{60}=0,1\ (mol)\\n_{C_2H_5OH}=\dfrac{6}{46}=0,13\ (mol)$

Phản ứng xảy ra:

$\rm CH_3COOH+C_2H_5OH\overset{H_2SO_4\ (d ac),\ t^°}\leftrightarrows CH_3COOC_2H_5+H_2O$

Vì: $\rm n_{CH_3COOH}<n_{C_2H_5OH}$ nên tính toán theo $\rm CH_3COOH$.

$\rm n_{CH_3COOC_2H_5}=n_{CH_3COOH}=0,1\ (mol)$

$\rm\to m=0,1×88×80\%=7,04\ (g)$