Molar mass of CO_2=44g/mol
No of moles=1.5mol
Mass of CO_2
[tex]\\ \sf\longmapsto No\:of\:moles(Molar\:mass)[/tex]
[tex]\\ \sf\longmapsto 44(1.5)[/tex]
[tex]\\ \sf\longmapsto 66g[/tex]
And
Molar mass of water=18g/mol
No of moles=2
Mass of H_2O
[tex]\\ \sf\longmapsto 2(18)=36g[/tex]
Total mass of products
[tex]\\ \sf\longmapsto 66+36=102g[/tex]
EMPIRICAL FORMULA:-CH_2O_3
Empirical formula mass:-
[tex]\\ \sf\longmapsto 12+2(1)+3(16)=12+2+48=62g[/tex]
We need n now
[tex]\\ \sf\longmapsto n=\dfrac{Molecular\:mass\:of\; products}{Empirical\: formula\:mass}[/tex]
[tex]\\ \sf\longmapsto n=\dfrac{102}{62}[/tex]
[tex]\\ \sf\longmapsto n=1.64\approx 2[/tex]
Now
[tex]\\ \sf\longmapsto Molecular\: formula=n(Empirical\: formula)[/tex]
[tex]\\ \sf\longmapsto 2(CH_2O_3)[/tex]
[tex]\\ \sf\longmapsto C_2H_4O_6[/tex]
It's Ethene Oxide .
Answer is Ethene(C_2H_4)
Option B is correct
Done
