Question

plzz explain the excess part in detail​

Answer 1

Moles of Sulphur = given mass/gram molecular mass

= 16/32 = 0.5 moles

Moles of Oxygen = given mass/gram molecular mass

= 100/32 = 3.125 moles

The chemical equation is

S + O₂ → SO₂

So, 1 mole of S reacts with 1 mole of O₂ to produce 1 mole of SO₂.

In the given case,

0.5 moles of S reacts with 0.5 moles of O₂

O₂ left over after the reaction is 3.125-0.5 = 2.625

Since 'S' is completely consumed in the reaction, it is termed as "Limiting reactant"

whereas 'O₂' is called "Excess reactant", as it is present in excess (present even after one of the reactant is completely comsumed)

Mass of remaining O₂ (Excess reactant) = moles x molecular mass

= 2.625 x 32 = 84 grams

Hope it helps!!