- Identify the stoichiometric ratio from the balanced equation \(2 KClO_3 \rightarrow 2 KCl + 3 O_2\), which is 2 moles of \(KClO_3\) to 3 moles of \(O_2\).
- Apply the given quantity of 2 moles of \(KClO_3\) to this ratio: \(2 \times \frac{3}{2} = 3\).
- Conclude that 3 moles of \(O_2\) are produced.
- The final answer is \(\boxed{3}\).
Explanation
- Analyzing the Chemical Reaction
We are given the chemical decomposition reaction for potassium chlorate:
\[2 KClO_3 \rightarrow 2 KCl + 3 O_2\]
Our goal is to determine how many moles of oxygen gas (\(O_2\)) are produced when \(2\) moles of potassium chlorate (\(KClO_3\)) decompose. The balanced chemical equation provides the molar ratio between the reactants and products. In this specific equation, the coefficients indicate that \(2\) moles of \(KClO_3\) react to produce \(2\) moles of \(KCl\) and \(3\) moles of \(O_2\).
- Identifying the Stoichiometric Ratio
The stoichiometric coefficients in the balanced equation tell us the ratio in which substances react and are produced. From the equation \(2 KClO_3 \rightarrow 2 KCl + 3 O_2\), we can clearly see the relationship between \(KClO_3\) and \(O_2\):
\[\frac{\text{moles of } O_2}{\text{moles of } KClO_3} = \frac{3}{2}\]
This ratio means that for every \(2\) moles of potassium chlorate that decompose, exactly \(3\) moles of oxygen gas are produced.
- Calculating the Amount of Oxygen Produced
Given that we start with exactly \(2\) moles of \(KClO_3\), we can use the ratio identified in the previous step:
\[\text{Moles of } O_2 = \text{moles of } KClO_3 \times \frac{3 \text{ moles } O_2}{2 \text{ moles } KClO_3}\]
Substituting the given value:
\[\text{Moles of } O_2 = 2 \times \frac{3}{2} = 3\]
Thus, the decomposition of \(2\) moles of potassium chlorate produces \(3\) moles of oxygen gas. Comparing this to the given options (1, 2, 3, 6), the correct answer is 3.
Examples
In industrial chemistry, understanding stoichiometry is essential for manufacturing processes. For instance, if a company produces oxygen from potassium chlorate for specialized breathing equipment, they need to know exactly how much raw material to input to get the required amount of gas. This calculation ensures that resources are not wasted and that the required production quotas are met accurately, demonstrating the importance of balancing equations and stoichiometric ratios in real-world engineering and manufacturing tasks.