Question

Consider the conversion of oxegen O2 to ozone O3.

3O2 -> 2O3
What is the theoretical yeild of O3 in grams from 305g of O2?
What is the percent yeild of the reaction actually gives u 111 g of O3?

Answer 1

The reaction has a 36.43% percent yield. This suggests that part of the reactants were not transformed into products and that the reaction did not proceed to its full potential.

What is moles ?

The mole, which is denoted by the symbol "mol," is the volume of a system that has the same number of atoms as there are in 0.012 kilo grammes of carbon 12.

The chemical equation for converting
`O_(2)` to
`O_(3)` is as follows:
`O_(3)`

3
`O_(2)` ⇒ 2
`O_(3)`

From above we can conclude that the equation that 3 moles of
`O_(2)` react to create 2 moles of
`O_(3)`.

By applying the the mole ratio and molar mass of
`O_(3)`,

the theoretical yield of O3 will be-

`O_(3)` mol mass = 3 x O mol mass = 3 x 16.00 g/mol = 48.00 g/mol.

o find the number of moles of O2, we can use the formula:

moles =
`(mass)/(molar mass)`

moles of
`O_(2)` =
`(305 g)/(32.00)` g/mol = 9.53 mol

What is theoretical yield ?

The theoretical yield of
`O_(3)` from the balanced equation using the following formula:

Moles of
`O_(3)` =
`(2)/(3)` x moles of
`O_(2)`

=
`(2)/(3)` x 9.53 mol = 6.35 mol

Mass of O3 = moles of
`O_(3)` x molar mass of
`O_(3)`

= 6.35 mol x 48.00 g/mol = 304.80 g

= 304.80 g.

Percent yield = (
`(actual yield)/( theoretical yield)`) x 100%

Given that the actual yield = 111 g

Percent yield = (
`(111 g )/(304.80 g)`) x 100%

= 36.43%