Кто решит тот молодец!

$2C_{6} H_{4} C H_{3} (OH)+2K-\ \textgreater \ 2C_{6} H_{4108} C H_{3} (OK)+H2$

$HOC H_{2} C H_{2}OH+2K-\ \textgreater \ KOC H_{2} C H_{2}OK+ H_{2}$

$n= \frac{V}{ V_{m} }$

$V_{m} =22,4$ дм3\моль

$n( H_{2}) = \frac{20,7}{22,4} =0,924$ моль

Пусть $n(C_{6} H_{4} C H_{3} (OH))= n_{1}$ моль, а $n(HOC H_{2} C H_{2}OH)=n_{2}$ моль

$0,5n(C_{6} H_{4} C H_{3} (OH))=n_{1}( H_{2} )=0,5n_{1}$ моль

$n(HOC H_{2} C H_{2}OH)=n_{2}(H_{2})=n_{2}$ моль

Составим систему из уравнений:

$\left \{ {{0,5n_{1}+n_{2}=0,924} \atop {108n_{1}+62n_{2}=61,2}} \right.$

$\left \{ {{n_{2}=0,924-0,5n_{1}} \atop {108n_{1}+62(0,924-0,5n_{1})=61,2}} \right.$

$\left \{ {{n_{2}=0,924-0,5n_{1}} \atop {108n_{1}+57,288-31n_{1}=61,2}} \right.$

$\left \{ {{{n_{2}=0,924-0,5n_{1}} \atop {77n_{1}=3,912}} \right.$

$\left \{ {{n_{2}=0,924-0,5n_{1}} \atop {n_{1}= \frac{3,912} {77}=0,0508 }} \right.$

$\left \{ {{n_{1}=0,0508} \atop {n_{2}=0,924-0,5*0,0508=0,8986}} \right.$

$w(C_{6} H_{4} C H_{3} (OH))= \frac{0,0508*108}{61,2} *100=8,965$ % ≈9%