Question

Someone explain it plz​

Answer 1

Molar mass of CO_2=44g/mol

No of moles=1.5mol

Mass of CO_2

`\\ \sf\longmapsto No\:of\:moles(Molar\:mass)`

`\\ \sf\longmapsto 44(1.5)`

`\\ \sf\longmapsto 66g`

And

Molar mass of water=18g/mol

No of moles=2

Mass of H_2O

`\\ \sf\longmapsto 2(18)=36g`

Total mass of products

`\\ \sf\longmapsto 66+36=102g`

EMPIRICAL FORMULA:-CH_2O_3

Empirical formula mass:-

`\\ \sf\longmapsto 12+2(1)+3(16)=12+2+48=62g`

We need n now

`\\ \sf\longmapsto n=(Molecular\:mass\:of\; products)/(Empirical\: formula\:mass)`

`\\ \sf\longmapsto n=(102)/(62)`

`\\ \sf\longmapsto n=1.64\approx 2`

Now

`\\ \sf\longmapsto Molecular\: formula=n(Empirical\: formula)`

`\\ \sf\longmapsto 2(CH_2O_3)`

`\\ \sf\longmapsto C_2H_4O_6`

It's Ethene Oxide .

Answer is Ethene(C_2H_4)

Option B is correct

Done