2 grams of sulphur is completely burnt in oxygen to form SO2. what is the volume of oxygen consumed at STP.

Answer 1

According to Avogadro's hypothesis, a mole of gas at STP has the same volume as any other gas at STP. This standard molar volume has the value of 22.414 Liters. For the combustion of sulphur in oxygen, the balanced equation is:

S(solid) + O₂ (g) → SO₂(g)

The equation tells us that 1 mole of Sulphur uses 1 mole of Oxygen to produce 1 mole of Sulphur Dioxide. Dividing 2.00 grams by the molecular weight of Sulphur (32.06), we find that we have 2.00/32.06 moles of Sulphur = 0.06238 moles. That means (becuase of the 1:1 proportions of reaction) 0.06238 moles of oxygen were consumed during the reaction. Now we can set up the proportion:

1 mole/22.4 liter = 0.06238 moles/X Liters

Using Algebra, and solving for X, we have: X Liters = 0.06238 moles/1 moles/ 22.4 L

X = 0.06238moles/22.4 moles per Liter = 2.78 x 10^-3 Liters, or 2.78 ml.