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Chapter 8: Problem 32

Balance this chemical equation by inspection: \(\mathrm{KClO}_{3} \rightarrow \mathrm{KCl}+\mathrm{O}_{2}\)

Short Answer

Expert verified
To balance the given chemical equation \(\mathrm{KClO}_{3} \rightarrow \mathrm{KCl}+\mathrm{O}_{2}\), first balance the oxygens by adding a coefficient of 3/2 to \(\mathrm{O}_{2}\), resulting in \(\mathrm{KClO}_{3} \rightarrow \mathrm{KCl}+\frac{3}{2}\mathrm{O}_{2}\). Then, convert fractional coefficients to whole numbers by multiplying all coefficients by the least common denominator, which is 2 in this case. The balanced equation becomes \(2\mathrm{KClO}_{3} \rightarrow 2\mathrm{KCl} + 3\mathrm{O}_{2}\).

Step by step solution

01

Identify atoms on both sides

On the left side of the equation, we have one molecule of potassium chlorate (\(\mathrm{KClO}_{3}\)), containing: - 1 atom of potassium (K) - 1 atom of chlorine (Cl) - 3 atoms of oxygen (O) On the right side of the equation, we have one molecule of potassium chloride (\(\mathrm{KCl}\)) and one molecule of oxygen gas (\(\mathrm{O}_{2}\)), containing: - 1 atom of potassium (K) - 1 atom of chlorine (Cl) - 2 atoms of oxygen (O) Step 2: Balance oxygens by adding a coefficient to the \(\mathrm{O}_{2}\) term
02

Balance oxygens

We currently have 3 oxygen atoms on the left side of the equation and 2 on the right side. To balance the oxygens, we need to add a coefficient to the oxygen gas term so that there are equal oxygen atoms on both sides. In this case, we will add coefficient 3/2 to \(\mathrm{O}_{2}\), now we have: \[\mathrm{KClO}_{3} \rightarrow \mathrm{KCl}+\frac{3}{2}\mathrm{O}_{2}\] Step 3: Convert fractional coefficients to whole numbers.
03

Convert fractional coefficients

It's preferred to have whole number coefficients in a chemical equation. To achieve this, we can multiply the equation by the least common denominator (LCD) of all the coefficients. In this case, the LCD of the coefficients is 2. Multiplying all coefficients by the LCD, we get: \[ 2\times\mathrm{KClO}_{3} \rightarrow 2\times\mathrm{KCl} + 2\times\frac{3}{2}\mathrm{O}_{2} \] Which simplifies to: \[ 2\mathrm{KClO}_{3} \rightarrow 2\mathrm{KCl} + 3\mathrm{O}_{2} \] Now we have a balanced equation with whole number coefficients.

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Most popular questions from this chapter

Write a net ionic equation for the precipitation reaction, if any, that occurs when aqueous solutions of the following ionic compounds are mixed: (a) \(\mathrm{Bi}\left(\mathrm{NO}_{3}\right)_{3}\) and \(\mathrm{NaOH}\) (b) \(\mathrm{K}_{2} \mathrm{SO}_{4}\) and \(\mathrm{SrI}_{2}\) (c) \(\mathrm{Cu}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\) and \(\mathrm{Na}_{3} \mathrm{PO}_{4}\)

Aqueous solutions of sodium sulfide and iron(III) nitrate are combined. (a) Does a precipitation reaction occur? Explain. (b) If it does, write a net ionic equation for the reaction.

Balance the following equation and classify the reaction it represents as combination, decomposition, single-replacement, or double-replacement: \(\mathrm{Fe}+\mathrm{CdCl}_{2} \rightarrow \mathrm{FeCl}_{3}+\mathrm{Cd}\)

Draw dot diagrams for the reactants and products in each equation, and indicate which bonds are broken and which bonds are formed: (a) \(\mathrm{CH}_{4}+\mathrm{I}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{I}+\mathrm{HI}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{F}_{2} \rightarrow \mathrm{FCH}_{2} \mathrm{CH}_{2} \mathrm{~F}\)

Under the proper conditions, hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}(l)\), can be converted to water and oxygen gas. Write a balanced equation for this reaction and assign it a reaction type.
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