$69)$

Từ sơ đồ phản ứng ta có:

$H\%^{(1)}H\%^{(2)}n_{napthalene}=n_{DEP}$

`<=>`$75\%85\%\dfrac{x}{128}=\dfrac{100}{222}$

`=>`$x =\dfrac{6400}{111\cdot \:75\%\cdot \:85\%}\approx 90.4(kg)$

Cũng từ sơ đồ phản ứng ta có:

$H\%^{(2)}\times n_{ethylene \,\, glycol}=2n_{DEP}$

`<=>`$85\% \dfrac{y}{62}=2\times \dfrac{100}{222}$

`=>`$y=\dfrac{6200}{111\times \:85\%} \approx 65.7(kg)$

`=>`$x+y=\dfrac{6400+6200\times \:75\%}{111\times \:75\%\times \:85\%}\approx 156.2(kg)$