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SpiritInly

I think it's b

let's work out empirical formula:

CO2 + H20 is CH2O3

the mass of CH2O3 is 12+2+(16×3)=62

work out mass of CO2

mass=number of moles × relative formula mass

mass=1.5 mol × 44 = 66

work out mass of H20

mass=number of moles × relative formula mass

mass=2 × 18 = 36

so the molecular formula is 66+36=102

to calculate simplest molecular formula:

molecular formula mass/empirical formula mass

102/62=2 (rounded to nearest whole number)

back to the empirical formula which is CH2O3

multiply each element by 2 which is

C2H4O6

answer:C2H4

not 100% sure though, but I hope it's been of some use to you:)
  • 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

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