$\rm n_{C_2H_5OH}=\dfrac{4,6}{46}=0,1(Mol)$
$\rm PTHH:C_2H_5OH+3O_2 \xrightarrow{t^o} 2CO_2+3H_2O$

$\rm a.n_{O_2}=3n_{C_2H_5OH}=0,3(mol)$

$\rm V_{O_2}=0,3.22,4=6,72(l)$

$\rm V_{kk}=6,72:20\%=33,6(l)$
$\rm b.V_{C_2H_5OH}=\dfrac{4,6}{0,8}=5,75(ml)$

$\rm V_{r 8^o}={5,85.100}{8}=73,125(ml)$
$\rm c.PTHH:C_2H_5OH+O_2 \xrightarrow{men giấm,t^o} CH_3COOH+H_2O$
$\rm n_{CH_3COOH}=n_{C_2H_5OH}=0,1(mol)$

$\rm H=60\%$

$\rm m_{CH_3COOH}=0,1.60.60\%=3,6(g)$

Đáp án:

$\rm a)\\n_{C_2H_5OH}=\dfrac{4,6}{46}=0,1\ (mol)$

Phương trình phản ứng:

$\rm C_2H_5OH+3O_2\xrightarrow{t^°}2CO_2+3H_2O$

Theo phương trình:

$\rm n_{O_2}=3n_{C_2H_5OH}=0,3\ (mol)$

$\rm V_{kk}=\dfrac{0,3×22,4}{20\%}=33,6\ (L)$

$\rm b)\\V_{C_2H_5OH}=\dfrac{4,6}{0,8}=5,75\ (mL)$

$\rm V_{ruou \ 8^°}=\dfrac{5,75}{8\%}=71,875\ (mL)$

$\rm c)$

Phương trình phản ứng:

$\rm C_2H_5OH+O_2\xrightarrow{men,\ t^°}CH_3COOH+H_2O$

Theo phương trình:

$\rm n_{CH_3COOH}=n_{C_2H_5OH}=0,1\ (mol)$

Khối lượng acetic acid thực tế thu được:

$\rm m_{CH_3COOH}=0,1×60×60\%=3,6\ (g)$