Chemistry Stoichiometry ETEA 2023
PMDC Verified Question 33 of 105
Oxygen can be prepared by the decomposition of potassium chlorate (\( \text{KClO}_3 \)). How many moles of oxygen \( \text{O}_{2(g)} \) can be formed by taking 12 moles of potassium chlorate (\( \text{KClO}_3 \)) according to the following equation?

$$ 2\text{KClO}_{3(s)} + \text{heat} \rightarrow 2\text{KCl}_{(s)} + 3\text{O}_{2(g)} $$
A
12 moles of oxygen
B
15 moles of oxygen
C
18 moles of oxygen
D
21 moles of oxygen
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: 18 moles of oxygen
Concept:

Using the molar ratios established by the balanced chemical equation, we can directly convert the moles of a decomposed reactant into the moles of generated product.

Formula:

$$ \frac{\text{Moles of O}_2}{\text{Coefficient of O}_2} = \frac{\text{Moles of KClO}_3}{\text{Coefficient of KClO}_3} $$

Solution:

  • The balanced equation establishes the molar ratio: 2 moles of \( \text{KClO}_3 \) decompose to produce exactly 3 moles of \( \text{O}_2 \).


  • We are provided with 12 moles of \( \text{KClO}_3 \).


  • Let \( x \) be the moles of \( \text{O}_2 \) produced. Set up a cross-multiplication:
    \( \frac{2}{12} = \frac{3}{x} \)


  • \( 2x = 36 \)


  • \( x = 18 \text{ moles of O}_2 \).


Why other options are incorrect:

  • 12 moles: Assumes an incorrect 1:1 molar ratio, ignoring the coefficients entirely.


  • 15 & 21 moles: Random arithmetic outputs that do not align with the strict 2:3 reaction stoichiometry.

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