Chemistry - Mole Concept Question with Solution | TestHub

$C_{57}{H}_{110}{O}_6+\frac{163}{2}{O}_2\to 57C{O}_2+55H_{2}O$

Molecular weight of$C_{57}{H}_{110}{O}_6=890$

Moles of$C_{57}{H}_{110}{O}_6=\frac{\mathrm{w}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}\left(\mathrm{g}\mathrm{m}\right)}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{ }\mathrm{w}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}$

$=\frac{445}{890}=\frac{1}{2}$

According to stoichiometry

$\because 1 \mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}$of$C_{57}{H}_{110}{O}_6$gives$=55 \mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}$of$H_{2}O$

$\because \frac{1}{2} \mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}$of$C_{57}{H}_{110}{O}_6$gives$=55\times \frac{1}{2} \mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}$of$H_{2}O$

Wt. of$H_{2}O=$mole of$H_{2}O\times M$Wt. of$H_{2}O$

$=\frac{1}{2}\times 55\times 18=495 gm$