Oxygen can be produced from the catalytic decomposition of kclo3 as shown in the balanced equation below. 2 kclo3 → 2 kcl + 3 o2 what is the percent yield if 3.20 grams of oxygen are formed from the reaction of 12.3 grams of kclo3?

the balanced equation for the decomposition of KClO₃ is as follows

2KClO₃ ---> 2KCl + 3O₂

stoichiometry of KClO₃ to O₂ is 2:3

mass of KClO₃ decomposed is - 12.3 g

number of moles of KClO₃ - 12.3 g / 122. 5 g/mol = 0.1004 mol

acccording to molar ratio

when 2 mol of KClO₃ decompose - 3 mol of O₂ is formed

therefore when 0.1004 mol of KClO₃ decompose - 3/2 x 0.1004 mol = 0.1506 mol

of O₂ is formed

mass of O₂ expected to be formed  - 0.1506 mol x 32 g/mol = 4.82 g

this is the theoretical yield - 4.82 g

percentage yield = (actual yield / theoretical yield) x 100 %

actual yield = 3.20 g

theoretical yield = 4.82 g

percentage yield = (3.20 g / 4.82 g ) x 100 % = 66.4 %

percentage yield = 66.4 %

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