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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?

Consider the conversion of oxegen O2 to ozone O3. 3O2 -> 2O3 What is the theoretical-example-1
User Tometoyou
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1 Answer

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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%

User CdkMoose
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