Ta có :

$n_{phenyl\ axetat} = n_{vinyl\ axetat} = n_{vinyl\ acrylat} = n_{triolein} = 1(mol)$

Ta có :

$n_{Br_2} = n_{vinyl\ axetat} + 2n_{vinyl\ acrylat} + 3n_{triloein} = 6(mol)$

$m_{Br_2} = 160.6 = 960(gam)$

Triolein t/d đực 3 mol Br2 mà em?

Hix, làm hết bài bạn đăng mình xỉu ớ :((

\(n_{H_2O}=\dfrac{12.6}{18}=0.7\left(mol\right)\)

\(n_{CO_2}=0.3\cdot3=0.9\left(mol\right)\)

\(m_{tăng}=m_{CO_2}+m_{H_2O}-m_{CaCO_3}=7.2\left(g\right)\)

\(m_{CaCO_3}=0.9\cdot44+0.7\cdot18-7.2=45\left(g\right)\)

Mk cám ơn bạn nhiều nha mấy câu ngày có nhiều câu mk ko chắc nên mới nhờ làm☺️

\(n_{CO_2}=0.7\left(mol\right)\)

\(\overline{c}=\dfrac{0.7}{0.25}=2.8\)

=> X là các este no, đơn chức, mạch hở

\(n_{CO_2}=n_{H_2O}=0.7\left(mol\right)\)

\(m=0.7\cdot\left(44+18\right)=43.4\left(g\right)\)

\(n_{C_4H_8O_2}=\dfrac{19.8}{88}=0.225\left(mol\right)\)

\(\Rightarrow n_{CO_2}=n_{H_2O}=0.225\cdot4=0.9\left(mol\right)\)

\(m_{\text{bình tăng}}=0.9\cdot44+0.9\cdot18=55.8\left(g\right)\)

\(m_{tăng}=m_{CO_2}+m_{H_2O}=15.5\left(g\right)\)

\(n_{CO_2}=n_{H_2O}=\dfrac{15.5}{44+18}=0.25\left(mol\right)\)

\(m_{Na_2CO_3}=0.25\cdot106=26.5\left(g\right)\)

\(n_{CO_2}=n_{H_2O}=\dfrac{30}{100}=0.3\left(mol\right)\)

\(m_{H_2O}=0.3\cdot18=5.4\left(g\right)\)

\(n_{H_2O}=\dfrac{5.4}{18}=0.3\left(mol\right)\)

\(n_{O_2}=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)

Bảo toàn O : 

\(n_{CO_2}=\dfrac{0.1\cdot2+0.45\cdot2-0.3}{2}=0.4\left(mol\right)\)

\(n_{CO_2}=n_{CaCO_3}=0.4\left(mol\right)\)

\(m_{CaCO_3}=0.4\cdot100=40\left(g\right)\)

\(n_{O_2}=\dfrac{1.904}{22.4}=0.085\left(mol\right)\)

Bảo toàn O : 

\(n_{CO_2}=n_{H_2O}=\dfrac{1.62+\dfrac{1.904}{22.4}\cdot32}{44+18}=0.07\left(mol\right)\)

\(n_{este}=\dfrac{0.07\cdot2+0.07-0.085\cdot2}{2}=0.02\left(mol\right)\)

\(M=\dfrac{1.62}{0.02}=81\)

\(CT:C_3H_6O_2,C_4H_8O_2\)

\(n_{CO_2}=n_{H_2O}\)

\(\Leftrightarrow\dfrac{m+2.8}{44}=\dfrac{m-2.4}{18}\)

\(\Leftrightarrow m=6\)

\(n_{CO_2}=n_{H_2O}=\dfrac{6+2.8}{44}=0.2\left(mol\right)\)

\(n_{este}=\dfrac{6-0.2\cdot12-0.2\cdot2}{32}=0.1\)

\(n=\dfrac{0.2}{0.1}=2\)

\(CT:C_2H_4O_2\)