Question

a common laboratory preparation of oxygen (o₂) gas is the thermal decomposition of potassium chlorate (kclo₃). assuming complete decomposition, calculate the number of grams of o₂ gas that can be obtained from 33.57 g kclo₃. (the products are kcl and o₂). round your answer to 4 significant digits.

Explanation:

Step1: Write the balanced chemical equation

The thermal decomposition of \(KClO_3\) is \(2KClO_3(s)\xrightarrow{\Delta}2KCl(s)+3O_2(g)\)

Step2: Calculate the molar mass of \(KClO_3\)

The molar mass of \(K\) is \(39.10\space g/mol\), \(Cl\) is \(35.45\space g/mol\), and \(O\) is \(16.00\space g/mol\).
\(M_{KClO_3}=39.10 + 35.45+(3\times16.00)=122.55\space g/mol\)

Step3: Calculate the number of moles of \(KClO_3\)

Using the formula \(n=\frac{m}{M}\), where \(m = 33.57\space g\) and \(M = 122.55\space g/mol\)
\(n_{KClO_3}=\frac{33.57\space g}{122.55\space g/mol}\approx0.274\space mol\)

Step4: Use the mole - ratio from the balanced equation

From \(2KClO_3(s)\xrightarrow{\Delta}2KCl(s)+3O_2(g)\), the mole - ratio of \(KClO_3\) to \(O_2\) is \(2:3\).
Let \(n_{O_2}\) be the number of moles of \(O_2\). Then \(\frac{n_{KClO_3}}{n_{O_2}}=\frac{2}{3}\), so \(n_{O_2}=\frac{3}{2}n_{KClO_3}\)
Substitute \(n_{KClO_3}= 0.274\space mol\) into the equation: \(n_{O_2}=\frac{3}{2}\times0.274\space mol = 0.411\space mol\)

Step5: Calculate the mass of \(O_2\)

The molar mass of \(O_2\) is \(M_{O_2}=32.00\space g/mol\)
Using the formula \(m = n\times M\), where \(n = 0.411\space mol\) and \(M = 32.00\space g/mol\)
\(m_{O_2}=0.411\space mol\times32.00\space g/mol = 13.152\space g\approx13.15\space g\)

Answer:

\(13.15\space g\)