1)m(р-ра KHCO3)=q*v=75*1.07=80.25(г.)
2)m(в-ва KHCO3)=w*m=80.25*0.1
3)ню общее(KHCO3)=m(KHCO3)/M(KHCO3)=8.025/100=0.08(моль)
4)пусть m(ch3-cooh)=x
,то m(c2h5-cooh)=5.64-x
5)ню(ch3-cooh)=x/60
ню(c2h5-cooh)=5.64-x
6)ню1(KHCO3)=y
7)ню2(KHCO3)=0.08-y
8 x/60=y
5.64-x/74=0.08-y
Решаем эту систему.
5.64-60y/74=0.08025-y
5.64-60y=5.94-74y
14y=0.3
y=0.043
x=0.043*60
x=2.6
w(ch3-cooh)=2.6/5,64=46%
w(c2h5-cooh)=100-w(ch3-cooh)=54%
Ответ:w(ch3-cooh)=46%;w(c2h5-cooh)=54%