Câu hỏi 12

\({m_{metylmetacrylat}} = {{{m_{PMM}}.100} \over {{H_{trung\,hop}}}} = {{120.100} \over {80}} = 150\,Kg =  > \,\,{n_{metylmetacrylat}} = 1,5.1{0^3}\,mol\)

Ta có:

\(\eqalign{& {n_{C{H_2} = C\left( {C{H_3}} \right) - COOH}} = {n_{C{H_3}OH}} = {{{n_{metylmetacrylat}}.100} \over H} = 2,5.1{0^3}\,mol & \cr & = > \,\,{n_{C{H_2} = C\left( {C{H_3}} \right) - COOH}} = 215.1{0^3}\,g = 215\,Kg,\,\,\,{n_{C{H_3}OH}} = 80.1{0^3}\,g = 80\,kg \cr} \)

Đáp án A