Question

2 kclo₃ → 2 kcl + 3 o₂

1. given 33.79 grams of o₂, how many moles of kclo₃ are produced?
2. given 2.8 moles of o₂, how many grams of kcl are also used?
3. given 6.25 moles of kclo₃, how many grams of kcl are produced?

Explanation:

Step1: Calculate moles of \(O_2\)

The molar mass of \(O_2\) is \(M = 32\space g/mol\).
The number of moles of \(O_2\), \(n_{O_2}=\frac{m}{M}=\frac{33.79\space g}{32\space g/mol}\approx1.056\space mol\)

Step2: Use mole ratio from the balanced equation

From the balanced equation \(2KClO_3
ightarrow2KCl + 3O_2\), the mole ratio of \(KClO_3\) to \(O_2\) is \(\frac{n_{KClO_3}}{n_{O_2}}=\frac{2}{3}\)
So \(n_{KClO_3}=\frac{2}{3}n_{O_2}\)
Substitute \(n_{O_2} = 1.056\space mol\)
\(n_{KClO_3}=\frac{2}{3}\times1.056\space mol = 0.704\space mol\)

Step3: For part b)

From the balanced equation, mole ratio of \(KCl\) to \(O_2\) is \(\frac{n_{KCl}}{n_{O_2}}=\frac{2}{3}\)
Given \(n_{O_2}=2.8\space mol\), then \(n_{KCl}=\frac{2}{3}\times2.8\space mol=\frac{5.6}{3}\space mol\)
Molar mass of \(KCl\) is \(M = 74.55\space g/mol\)
\(m_{KCl}=n_{KCl}\times M=\frac{5.6}{3}\space mol\times74.55\space g/mol\approx139.14\space g\)

Step4: For part c)

From the balanced equation, mole ratio of \(KCl\) to \(KClO_3\) is \(1:1\)
Given \(n_{KClO_3} = 6.25\space mol\), then \(n_{KCl}=6.25\space mol\)
\(m_{KCl}=n_{KCl}\times M=6.25\space mol\times74.55\space g/mol = 465.94\space g\)

Answer:

a) \(0.704\space mol\)
b) \(139.14\space g\)
c) \(465.94\space g\)