Bài 48- Luyện tập- Rượu etylic, axit axetic và chất béo

Giải bài 1 trang 148 SGK Hóa 9

Bài 1: Cho các chất sau: rượu etylic, axit axetic, chất béo. Hỏi:

a) Phân tử chất nào có nhóm – OH? Nhóm – COOH?

b) Chất nào tác dụng với K? với Zn? Với NaOH? Với K2CO3?

Viết các phương trình hóa học

Lời giải:

a) Chất có nhóm – OH: rượu etylic (C2H5OH), CH3COOH.

Chất có nhóm – COOH: axit axetic (CH3COOH).

b) – Chất tác dụng được với K: rượu etylic và axit axetic:

2C2H5OH + 2K → 2C2H5OK + H2

2CH3COOH + 2K → 2CH3COOK + H2

– Chất tác dụng được với Zn: CH3COOH

2CH3COOH + Zn → (CH3COO)2Zn + H2

– Chất tác dụng được với NaOH : axit axetic và chất béo :

CH3COOH + NaOH → CH3COONa + H2O

(RCOO)3C3H5 + 3NaOH → C3H5(OH)3 + 3RCOONa

– Chất tác dụng với K2CO3: CH3COOH

2CH3COOH + K2CO3 → 2CH3COOK + CO2↑ + H2O.

Giải bài 2 trang 148 SGK Hóa 9

Bài 2: Tương tự chất béo, etyl axetat cũng có phản ứng thủy phân trong dung dịch axit và dung dịch kiềm. Hãy viết phương trình hóa học xảy ra khi đung etyl axetat với dung dịch HCl, dung dịch NaOH.

Lời giải:

Phản ứng của etyl axtat với dung dịch HCl:

CH3COOC2H5 + H2O –HCl,to->  CH3COOH + C2H5OH

Phản ứng của etyl axtat với dung dịch NaOH.

CH3COOC2H5 + NaOH −to→ CH3COONa + C2H5OH.

Giải bài 3 trang 149 SGK Hóa 9

Bài 3: Hãy chọn các chất thích hợp điền vào các dấu hỏi rồi hoàn thành các phương trình hóa học sau:

a) C2H5OH + ? → ? + H2

b) C2H5OH + ? → CO2 + ?

c) CH3COOH + ? → CH3COOK + ?

d) CH3COOH + ? → CH3COOC2H5 + ?

e) CH3COOH + ? → ? + CO2 + ?

g) CH3COOH + ? → ? + H2

h) Chất béo + ? → ? + muối của các axit béo.

Lời giải:

a) 2C2H5OH + 2Na → 2C2H5ONa + H2

b) C2H5OH + 3O2 −to→ 2CO2 + 3H2O.

c) 2CH3COOH + 2K → 2CH3COOK + H2.

d) CH3COOH + C2H5OH H2SO4, đn, to⇆ CH3COOC2H5 + H2O

e) 2CH3COOH + CaCO3 → (CH3COO)2Ca + CO2 + H2O.

g) 2CH3COOH + Mg → (CH3COO)2Mg + H2

h) Chất béo + Natri hidroxit → Glixerol + muối của các axit béo.

Giải bài 4 trang 149 SGK Hóa 9

Bài 4: Có ba lọ không nhãn đựng ba chất lỏng là: rượu etylic, axit axetic và dầu ăn tan trong rượu etylic. Chỉ dùng nước và quỳ tím, hãy phân biệt các chất lỏng trên.

Lời giải:

Trích mẫu thử và đánh số thứ tự:

– Lần lượt nhúng quỳ tím vào 3 mẫu thử trên

    + Mẫu làm quỳ tím hóa đỏ là axit axetic.

    + 2 mẫu còn lại không có hiện tượng gì.

– Cho hai chất lỏng còn lại cho vào nước, chất nào tan hoàn toàn đó là rượu etylic, còn lại là hỗn hợp dầu ăn tan trong rượu etylic.

Giải bài 5 trang 149 SGK Hóa 9

Bài 5: Khi xác định công thức của các chất hữu cơ A và B, người ta thấy công thức phân tử của A là C2H6O, còn công thức phân tử của B là C2H4O2. Để chứng minh A là rượu etylic, B là axit axetic cần phải làm thêm những thí nghiệm nào? Viết phương trình hóa học minh họa (nếu có).

Lời giải:

PTHH: 2C2H5OH + 2Na → 2C2H5ONa + H2↑.

– Ứng với CTPT: C2H6O, ta sẽ có 2 công thức cấu tạo

CH3 – O – CH3 và CH3CH2 – OH.

Cho A tác dụng bới Na nếu có khí bay ra thì đó là rượu etylic.

– Ứng với CTPT: C2H4O2 ta có các CTCT sau:

bai 48 luyen tap ruou 3

Cho B tác dụng với Na2CO3 nếu có khí thoát ra, chứng tỏ B là axit axetic.

2CH3COOH + Na2CO3 → 2CH3COONa + H2O + CO2

Giải bài 6 trang 149 SGK Hóa 9

Bài 6: Khi lên men dung dịch loãng của rượu etylic, người ta được giấm ăn.

a) Từ 10 lít rượu 8o có thể điều chế được bao nhiêu gam axit axetic? Biết hiệu xuất quá trình lên men là 92% và rượu etylic có D = 0,8 g/cm3.

b) Nếu pha khối lượng axit axetic trên thành dung dịch giấm 4% thì khối lượng dung dịch giấm thu được là bao nhiêu?

Lời giải:

a) Trong 10 lít rượu 80 thì có 10.8/100 = 0,8 lít rượu etylic

⇒ m rượu = V.D = 0,8.0,8.1000 = 640 (g)

Phương trình phản ứng lên men rượu:

C2H5OH + O2 → CH3COOH + H2O

bai 48 luyen tap ruou 1 1

Giải bài 7 trang 149 SGK Hóa 9

Bài 7: Cho 100g dung dịch CH3COOH 12% tác dụng vừa đủ với dung dịch NaHCO3 8,4%.

a) Hãy tính khối lượng dung dịch NaHCO3 đã dùng.

b) Hãy tính nồng độ phần trăm của dung dịch muối thu được sau phản ứng.

Lời giải:

PTHH:

bai 7 trang 149 sgk hoa 9 1 1
bai 7 trang 149 sgk hoa 9 2 1

Theo pt: nNaHCO3 = nCH3COOH = 0,2 (mol) → mNaHCO3 = 0,2. 84 = 16,8 (g)

bai 7 trang 149 sgk hoa 9 3 1

b) Theo pt: nCH3COONa = nCO2 = nCH3COOH = 0,2 (mol)

mCH3COONa = 82. 0,2 = 16,4 g

mCO2 = 0,2. 44 = 8,8 (g)

mdd = mCH3COOH + mNaHCO3 – mCO2 = 100 + 200 – 8,8 = 291,2 (g)

bai 48 luyen tap ruou 2 1

Lý thuyết Hóa học 9 Bài 48: Luyện tập: Rượu etylic, axit axetic và chất béo

ly thuyet bai 48 luyen tap 1

Bài tập Trắc nghiệm Hóa học 9 Bài 48 (có đáp án): Luyện tập: Rượu etylic, axit axetic và chất béo

Câu 1: Este là sản phẩm của phản ứng giữa

 A. axit và rượu.

 B. rượu và gluxit.

 C. axit và muối.

 D. rượu và muối.

Hiển thị đáp án

Đáp án: A

Câu 2: Hòa tan 30 ml rượu etylic nguyên chất vào 120 ml nước cất thu được

 A. rượu etylic có độ rượu là 20°.

 B. rượu etylic có độ rượu là 25°.

 C. rượu etylic có độ rượu là 30°.

 D. rượu etylic có độ rượu là 35°.

Hiển thị đáp án

Đáp án: A

bai tap bai 48 luyen tap ruou etylic axit axetic va chat beo a01 1

Câu 3: Cho các chất sau : Mg, Cu, CuO, NaCl, C2H5OH, Ba(OH)2. Số chất tác dụng được với dung dịch axit axetic là

 A. 3.

 B. 4.

 C. 5.

 D. 6.

Hiển thị đáp án

Đáp án: B

Các chất tác dụng được với axit axetic là: Mg; CuO; C2H5OH; Ba(OH)2.

Câu 4: Có ba lọ không nhãn đựng : rượu etylic, axit axetic, dầu vừng. Có thể phân biệt bằng cách nào sau đây ?

 A. Dùng quỳ tím và nước.

 B. Khí cacbon đioxit và nước.

 C. Kim loại kali và nước.

 D. Phenolphtalein và nước.

Hiển thị đáp án

Đáp án: A

– Sử dụng quỳ tím:

 + Quỳ tím chuyển sang màu đỏ → axit axetic.

 + Quỳ tím không đổi màu: rượu etylic, dầu vừng (nhóm I).

– Cho từng chất ở nhóm I vào nước:

 + Thu được dung dịch đồng nhất → rượu etylic.

 + Dung dịch tách thành hai lớp → dầu vừng.

Câu 5: Một chất hữu cơ A có khối lượng phân tử là 46 đvC. Công thức phân tử của A là

 A. C3H6O.

 B. C2H6O.

 C. C2H4O2.

 D. CH2O.

Hiển thị đáp án

Đáp án: B

Câu 6: Giấm ăn là

 A. dung dịch rượu etylic có nồng độ trên 10 %.

 B. dung dịch rượu etylic có nồng độ dưới 2 %.

 C. dung dịch axit axetic có nồng độ từ 2% – 5%.

 D. dung dịch axit axetic có nồng độ từ 5% – 10%.

Hiển thị đáp án

Đáp án: C

Câu 7: Hòa tan 15 gam CaCO3 vào dung dịch CH3COOH dư. Thể tích CO2 thoát ra ( đktc) là

 A. 2,24 lít.

 B. 3,36 lít.

 C. 4,48 lít.

 D. 5,60 lít.

Hiển thị đáp án

Đáp án: B

bai tap bai 48 luyen tap ruou etylic axit axetic va chat beo 1 1

Câu 8: Đun nóng chất béo với nước thu được

 A. glixerol và muối của một axit béo.

 B. glixerol và các axit béo.

 C. glixerol và axit hữu cơ.

 D. glixerol và muối của các axit béo

Hiển thị đáp án

Đáp án: B

Câu 9: Cho 13,6 gam hỗn hợp X gồm axit axetic và rượu etylic được trộn theo tỉ lệ mol tương ứng là 3 : 2. Đun nóng X với H2SO4 đặc một thời gian thu được m gam este CH3COOCH2CH3 với hiệu suất phản ứng là 80%. Giá trị của m là

 A. 7,04g.

 B. 8,80g.

 C. 10,56g.

 D. 11,00g.

Hiển thị đáp án

Đáp án: A

Gọi số mol của CH3COOH là 3a mol → số mol của CH3CH2OH là 2a mol

mX = 13,6 gam → 60.3a + 46.2a = 13,6 → a = 0,05 mol

PTHH:

bai tap bai 48 luyen tap ruou etylic axit axetic va chat beo 2 1

Giả sử hiệu suất là 100% thì rượu etylic hết, nên số mol sản phẩm phản ứng tính theo số mol rượu etylic.

Số mol este là: neste thực tế = neste lý thuyết. H = 0,1.80% = 0,08 mol

→ m = n.M = 0,08.88 = 7,04 gam.

Câu 10: Lên men 1 lít ancol etylic 23 độ thu được giấm ăn. Biết hiệu suất lên men là 100% và khối lượng riêng của ancol etylic là 0,8 gam/ml. Khối lượng axit axetic trong giấm là bao nhiêu?

 A. 240 gam.

 B. 230 gam.

 C. 480 gam.

 D. 460 gam.

Hiển thị đáp án

Đáp án: A

Vrượu = Độ rượu.Vdd = 0,23.1000 = 230 ml

→ mrượu = Vrượu.D = 230.0,8 = 184 gam

bai tap bai 48 luyen tap ruou etylic axit axetic va chat beo 3 1

Bản quyền bài viết thuộc trường trung học phổ thông Sóc Trăng. Mọi hành vi sao chép đều là gian lận.
Nguồn chia sẻ: Trường THPT Thành Phố Sóc Trăng (thptsoctrang.edu.vn)

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  43. Wow that was odd. I just wrote an really long comment but after I clicked submit my comment didn’t appear. Grrrr… well I’m not writing all that over again. Anyhow, just wanted to say superb blog!

  44. Can I simply just say what a comfort to discover an individual who genuinely knows what they’re talking about on the web. You definitely understand how to bring a problem to light and make it important. More and more people ought to look at this and understand this side of your story. I was surprised that you are not more popular since you most certainly have the gift.

  45. May I simply just say what a comfort to uncover someone that really understands what they are talking about on the web. You actually know how to bring a problem to light and make it important. More people must check this out and understand this side of your story. I can’t believe you’re not more popular since you most certainly have the gift.

  46. Excellent goods from you, man. I have be mindful your stuff previous to and you are just too great. I actually like what you have bought here, certainly like what you’re stating and the way by which you assert it. You make it entertaining and you still care for to keep it wise. I can’t wait to learn much more from you. That is really a wonderful website.

  47. Can I just say what a relief to discover somebody that genuinely understands what they are talking about on the web. You certainly understand how to bring a problem to light and make it important. More and more people ought to read this and understand this side of your story. I was surprised that you are not more popular given that you definitely have the gift.

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