0.2 g of S is reacted with excess of oxygen then determine the volume of sulpusu dioxide in the reaction.

S₈+ O₂ ----> SO2

Open in App

Solution

S8 + 8O2 = 8SO2

Always remember that " 1 mole of any gas at STP occupies a volume of 22.4 L"

weight of S8 is 256.5 AND Molecular weight of "S" is 32)

Given,
.2g of S8 is reacted

No.of moles=given mass/ molecular mass
no.of moles=.2/256.5
.
Vol. of S8=22.4×(.2/256.5)
=0.01746 L

Suggest Corrections

Similar questions

{2 S O}_{2} (g) + O_{2} (g) + {2 H}_{2} O (l) ⟶ {2 H}_{2} {S O}_{4}

5.6

S O_{2}

4.8

O_{2}

H_{2} {S O}_{4}

1 l i t

3 l i t

S O_{2}

S T P

S O_{3}

C_{3} H_{8}

560 ml of carbon monoxide is mixed with 500 ml of oxygen and ignited. The chemical equation for the reaction is as follows:

$2 C O + O_{2} \to 2 C O_{2}$

Calculate the volume of oxygen used and carbon dioxide formed in the above reaction.

Join BYJU'S Learning Program