Bài tập Hóa Hữu Cơ Hay phần Este

By Trang Ánh Nam -

 Hỗn hợp E gồm 4 este đều có công thức C8H8O2 và đều chứa vòng benzen. Cho 16,32 gam E tác dụng tối đa với V lít NaOH 1M (đung nóng) thu được 3,88 gam ancol và 18,78 gam hỗn hớp muối. Giá Trị của V là:

    A.190               B.100            C.120                  D.240  



Do mình chưa đủ trình nên phải chia ra làm 2 giai đoạn cho dể hiểu, tới khi nhập đạo thành Cao Thủ rồi thì ra thẳng luôn cũng được.

ü Giai đoạn 1: NaOH phản ứng với góc este không phản ứng với góc Phenol.

=>mgiữa=meste+mNaOH=16,32+0,12.40=21,12 gam.

ü Giai đoạn 2: NaOH tiếp tục phản ứng với góc Phenol.

NaOH + -OH à-ONa+H2O

a…………..a……….a……….a

ta thấy tổng khối lượng của rượu và muối là: 3,88+18,78=22,66 gam, lớn hơn khối lượng ở giữa nên áp dụng m tăng ta được.

mtăng=mvào-mra

ó22,66-21,12=ma.mol.NaOH-mH2O

ó1,54=a.40-a.18

=>a=0,07

Vậy tổng số mol NaOH phản ứng là: 0,12+0,07=0,19 mol.

=>V=190 đáp án A.

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