a)\(2CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+H_2\uparrow\)
b)\(n_{H_2}=\dfrac{1}{2}n_{CH_3COOH}\Rightarrow V_{H_2}=\dfrac{1}{2}V_{CH_3COOh}=\dfrac{1}{2}\cdot100=50ml\)
c)\(CH_3COOH+C_2H_5OH\xrightarrow[xtH_2SO_4đặc]{140^oC}CH_3COOC_2H_5+H_2O\)
\(n_{este}=n_{CH_3COOC_2H_5}=\dfrac{130,68}{88}=1,485mol\)
\(\Rightarrow n_{CH_3COOH}=1,485mol\Rightarrow m_{líthuyết}=1,485\cdot60=89,1g\)
\(m_{thựctế}=V_{CH_3COOH}\cdot D_{CH_3COOH}=100\cdot1,05=105g\)
Hiệu suất phản ứng:
\(H=\dfrac{m_{líthuyết}}{m_{thựctế}}\cdot100\%=\dfrac{89,1}{105}\cdot100\%=84,86\%\)
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