contestada

The industrial preparation of ozone (03) uses oxygen gas (O2), as shown in the equation below. 302(8) electricity 203(g) A If 96.0 kg of O2 gas actually yields 11.5 kg of 03. what is the percent yield in this reaction? A. 3.10% B. 8.35% C. 12.0% D. 18.0% ​

Respuesta :

The percent yield of the industrial preparation of ozone is the 12%.

How do we calculate the percent yield?

Percent yield of any reaction will be calculated by using the below equation:

% yield = (Actual yield / Theoretical yield) × 100%

Given chemical reaction is:

3O₂(g) → 2O₃(g)

Moles of oxygen will be calculated as:

n = W/M, where

  • W = given mass = 96kg = 96,000g
  • M = molar mass = 32g/mol

n = 96,000/32 = 3000mol

From the stoichiometry of the reaction it is clear that:

3 moles of oxygen = produces 2 moles of ozone

3000 moles of oxygen = produces 2/3(3000)=2000 moles of ozone

Mass of 2000 moles of ozone = (2000mol)(48g/mol) = 96,000g = 96kg

Actual yield of ozone = 11.5kg (given)

Now percent yield will be calculated as:

% yield = (11.5 / 96) × 100% = 11.97% = 12%

Hence required percent yield is 12%.

To know more about percent yield, visit the below link:

https://brainly.com/question/20349874

#SPJ1

Otras preguntas