Đáp án:

$\rm a)$

Phương trình phản ứng:

$\rm CH_3COOH+NaHCO_3\to CH_3COONa+CO_2+H_2O$

$\rm b)\\n_{CO_2}=\dfrac{6,72}{22,4}=0,3\ (mol)$

Theo phương trình:

$\rm n_{CH_3COOH}=n_{CO_2}=0,3\ (mol)$

$\rm a=m_{dd\ CH_3COOH}=\dfrac{0,3×60}{15\%}=120\ (g)$

$\rm c)\\m_{C_2H_5OH}=46×25\%×0,8=9,2\ (g)\\n_{C_2H_5OH}=\dfrac{9,2}{46}=0,2\ (mol)$

Phương trình phản ứng:

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

Vì $\rm n_{CH_3COOH}>n_{C_2H_5OH}\to$ Tính theo $\rm C_2H_5OH$.

$\rm n_{CH_3COOC_2H_5}=n_{C_2H_5OH}=0,2\ (mol)$

$\rm m_{CH_3COOC_2H_5}=0,2×88=17,6\ (g)$

 Ta có: 𝑛𝐶𝐻3𝐶𝑂𝑂𝐻=100.15%60=0,25(𝑚𝑜𝑙)nCH3​COOH​=60100.15%​=0,25(mol)

PT: 2𝐶𝐻3𝐶𝑂𝑂𝐻+𝐾2𝐶𝑂3→2𝐶𝐻3𝐶𝑂𝑂𝐾+𝐶𝑂2+𝐻2𝑂2CH3​COOH+K2​CO3​→2CH3​COOK+CO2​+H2​O

___________0,25_____0,125_________0,25____0,125 (mol)

⇒𝑚𝑑𝑑𝐾2𝐶𝑂3=0,125.13813,8%=125(𝑔)⇒mddK2​CO3​​=13,8%0,125.138​=125(g)

b, m dd sau pư = 100 + 125 - 0,125.44 = 219,5 (g)

⇒𝐶%𝐶𝐻3𝐶𝑂𝑂𝐾=0,25.98219,5.100%≈11,61%⇒C%CH3​COOK​=219,50,25.98​.100%≈11,61%

c, 𝐶2𝐻5𝑂𝐻+𝑂2𝑚𝑒𝑛𝑔𝑖𝑎𝑚→𝐶𝐻3𝐶𝑂𝑂𝐻+𝐻2𝑂C2​H5​OH+O2​mengiam​CH3​COOH+H2​O

𝑛𝐶2𝐻5𝑂𝐻(𝐿𝑇)=𝑛𝐶𝐻3𝐶𝑂𝑂𝐻=0,25(𝑚𝑜𝑙)nC2​H5​OH(LT)​=nCH3​COOH​=0,25(mol)

Mà: H = 90%

⇒𝑛𝐶2𝐻5𝑂𝐻(𝑇𝑇)=0,2590%=518(𝑚𝑜𝑙)⇒nC2​H5​OH(TT)​=90%0,25​=185​(mol)

⇒𝑚𝐶2𝐻5𝑂𝐻=518.46=1159(𝑔)⇒mC2​H5​OH​=185​.46=9115​(g)

⇒𝑉𝐶2𝐻5𝑂𝐻=11590,8=15,97(𝑚𝑙)⇒VC2​H5​OH​=0,89115​​=15,97(ml)

⇒𝑉𝑟ượ𝑢10𝑜=15,9710%=159,7(𝑚𝑙)⇒Vrượu10o​=10%15,97​=159,7(ml)10:00/-strong/-heart:>:o:-((:-hĐã gửiXem trước khi gửiThả Files vào đây để xem lại trước khi gửi

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