Question

when heated, kclo₃ decomposes into kcl and o₂. 2kclo₃ → 2kcl + 3o₂. if this reaction produced 22.8 g kcl, how many grams of o₂ were produced?

Explanation:

Step1: Calculate molar mass of KCl

The molar mass of K (potassium) is approximately 39.1 g/mol and Cl (chlorine) is approximately 35.45 g/mol. So molar - mass of KCl, $M_{KCl}=39.1 + 35.45=74.55$ g/mol.

Step2: Calculate moles of KCl

Given mass of KCl is $m_{KCl}=22.8$ g. Using the formula $n=\frac{m}{M}$, the number of moles of KCl, $n_{KCl}=\frac{22.8}{74.55}\approx0.306$ mol.

Step3: Determine mole - ratio of KCl and O₂

From the balanced chemical equation $2KClO_{3}
ightarrow2KCl + 3O_{2}$, the mole - ratio of KCl to O₂ is 2:3. So if $n_{KCl}$ moles of KCl are produced, the number of moles of O₂, $n_{O_{2}}=\frac{3}{2}n_{KCl}$.

Step4: Calculate moles of O₂

$n_{O_{2}}=\frac{3}{2}\times0.306 = 0.459$ mol.

Step5: Calculate molar mass of O₂

The molar mass of O₂ is $M_{O_{2}} = 2\times16=32$ g/mol.

Step6: Calculate mass of O₂

Using the formula $m = n\times M$, the mass of O₂, $m_{O_{2}}=n_{O_{2}}\times M_{O_{2}}=0.459\times32 = 14.688\approx14.7$ g.

Answer:

14.7 g