Verified answer

H-COOH+NaOH=H-COONa+H2O

CH3-COOH=NaOH=CH3-COONa+H2O

m(NaOH)=200*0,04=8;n(NaOH)=8:40=0,2;

Пусть в 1-ой р-ции затратилось "х" моль щелочи ,а во 2-ой "у",тогда х+у=0,2

n(H-COOH)=х моль;n(CH3-COOH)=y моль;m(H-COOH)=46x;m(CH3COOH)=60y;

Составляем систему и находим "Х" и" У"

                             х+у=0,2

                           46х+60у=10,6

                             у=0,1;х=0,1;     m(H-COOH)=0,1*46=4,6;

w(H-COOH)=4,6:10,6=0,43(43%);w(CH3-COOH)=57%