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Sagot :
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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