Question

2 kclo₃ → 2 kcl + 3 o₂

1. given 0.34 moles of o₂, how many moles of kclo₃ are used?
2. given 11 moles of o₂, how many moles of kcl are also produced?
3. given 6.45 moles of kclo₃, how many moles of kcl are produced?

Explanation:

Step1: Analyze the mole ratio

From the balanced chemical equation \(2\text{KClO}_3
ightarrow2\text{KCl}+3\text{O}_2\), the mole ratio of \(\text{KClO}_3\) to \(\text{O}_2\) is \(2:3\).

Step2: Set up the proportion

Let \(x\) be the moles of \(\text{KClO}_3\). Using the mole ratio \(\frac{x}{0.34}=\frac{2}{3}\).

Step3: Solve for \(x\)

Cross - multiply: \(3x = 2\times0.34\), so \(x=\frac{2\times0.34}{3}\approx0.23\) moles.

Step4: For the second part (question 5)

Mole ratio of \(\text{KCl}\) to \(\text{O}_2\) is \(2:3\). Let \(y\) be moles of \(\text{KCl}\). \(\frac{y}{11}=\frac{2}{3}\), cross - multiply \(3y = 2\times11\), \(y=\frac{22}{3}\approx7.33\) moles.

Step5: For the third part (question 6)

Mole ratio of \(\text{KClO}_3\) to \(\text{KCl}\) is \(2:2 = 1:1\). So if there are \(6.45\) moles of \(\text{KClO}_3\), moles of \(\text{KCl}\) is also \(6.45\) moles.

Answer:

1. Approximately \(0.23\) moles of \(\text{KClO}_3\) are used.
2. Approximately \(7.33\) moles of \(\text{KCl}\) are produced.
3. \(6.45\) moles of \(\text{KCl}\) are produced.