# What is the excess reagent and how many moles of it remain after 4.5 moles of sulphur react with 3.2 moles of ozone? 3S+2O_3→3SO_2

Jun 25, 2018

Ozone is the excess reagent, with $0.2$ moles left over.

#### Explanation:

The "excess reagent" is the one left over after all of the other one has been consumed in the reaction. In this reaction the balanced ratio is 3:2, so if the molar proportions are the same, they would both be use up. Whichever one has a molar proportion higher than this ratio is the excess reagent.

The ratio given is 4.5:3.2. Putting them into fractional form for comparison we have:

4.5/3.2 =? 3/2 The stoichiometric ratio is 1.5x S for every ${O}_{3}$, so we would need $1.5 \times 3.2 = 5$ moles of S. 4.5 is less than 5, so the ozone is in excess.

To find the amount left over, we use the same ratios to see how much ozone is actually consumed.

$\frac{4.5}{O} _ 3 = \frac{3}{2}$ ; ${O}_{3} = 4.5 \times \frac{2}{3} = 3.0$ consumed.

That leaves us with $3.2 - 3.0 = 0.2$ Moles of ${O}_{3}$ left over.