Question

When 20 moles of so2 and 15 moles of o2 are passed over catalyst . 10 moles of so3 are formed . The volume of resulting gaseous mixture is?

Answer 1

Answer:

10000cm3

Explanation:

This is a question on moles

To find the volume of the product sulphur six oxide we must first write a balanced chemical equation to determine the ratio within which the reactants react

The chemical equation is

2SO2 + O2 – 2SO3

We the proceed to dividing the moles of oxygen and sulphur dioxide with their corresponding stoichiometric numbers to determine which one of them is the limiting factor

20 ÷ 2 = 10 moles

15 ÷ 1 = 15 moles

Sulphur dioxide is the limiting factor

We have the moles of sulphur trioxide

We use the basic knowledge that one mole of any substance is equal to a thousand cubic centimetres

Therefore we substitute

If 1. =1000cm3

10moles

The answer is 10000cm3

Answer 2

Explanation:

10000 cm cube

10000 cm cubehope this helps you