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

$\\ \sf\longmapsto n=\dfrac{Molecular\:mass\:of\; products}{Empirical\: formula\:mass}$

$\\ \sf\longmapsto n=\dfrac{102}{62}$

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

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

Answer 2

Answer:

1). 1 mole of Carbon burnt in air

C + O2 →CO2

1 mole of carbon produces 1 mole of CO2 which is 44g of CO2

2). 1 mole of carbon is burnt in 16g of dioxygen

32g of O2 = 44g of CO2

1g of O2 = 44/32

CO2 (Dioxygen is limiting reagent)

16g of O2 = 4/32 × 16 = 22g of CO2 in one mole

3) 2 moles of Carbon burnt in 16g of dioxygen

16g of dioxygen is available, and thus it can combine with 0.5 mol of carbon to give 22g of CO2