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Chapter 3: Problem 91

A common demonstration in chemistry courses involves adding a tiny speck of manganese(IV) oxide to a concentrated hydrogen peroxide \(\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)\) solution. Hydrogen peroxide decomposes quite spectacularly under these conditions to produce oxygen gas and steam (water vapor). Manganese(IV) oxide is a catalyst for the decomposition of hydrogen peroxide and is not consumed in the reaction. Write the balanced equation for the decomposition reaction of hydrogen peroxide.

Short Answer

Expert verified
The balanced equation for the decomposition reaction of hydrogen peroxide is: \[2\,\mathrm{H}_2\mathrm{O}_2 \rightarrow 2\,\mathrm{O}_2 + 2\,\mathrm{H}_2\mathrm{O}\]

Step by step solution

01

1. Write the unbalanced equation

Write the unbalanced equation for the decomposition reaction of hydrogen peroxide: \[\mathrm{H}_2\mathrm{O}_2 \rightarrow \mathrm{O}_2 + \mathrm{H}_2\mathrm{O}\]
02

2. Balance the equation

In the unbalanced equation, we have 2 oxygen atoms on the left side and 3 on the right side. Also, there are 2 hydrogen atoms on the left side and 2 on the right. To balance the equation, we need to adjust the coefficients of the molecules. Multiply the water molecule by 2 to balance the number of hydrogen and oxygen atoms: \[2\,\mathrm{H}_2\mathrm{O}_2 \rightarrow 2\,\mathrm{O}_2 + 2\,\mathrm{H}_2\mathrm{O}\] Now we have 4 oxygen atoms and 4 hydrogen atoms on both sides of the equation, so it is balanced.
03

3. Write the balanced equation

The balanced equation for the decomposition reaction of hydrogen peroxide is: \[2\,\mathrm{H}_2\mathrm{O}_2 \rightarrow 2\,\mathrm{O}_2 + 2\,\mathrm{H}_2\mathrm{O}\] This equation shows the conversion of hydrogen peroxide into oxygen gas and water vapor, with the catalyst manganese(IV) oxide not appearing in the balanced equation.

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Most popular questions from this chapter

With the advent of techniques such as scanning tunneling microscopy, it is now possible to "write" with individual atoms by manipulating and arranging atoms on an atomic surface. a. If an image is prepared by manipulating iron atoms and their total mass is \(1.05 \times 10^{-20} \mathrm{~g}\), what number of iron atoms were used? b. If the image is prepared on a platinum surface that is exactly 20 platinum atoms high and 14 platinum atoms wide, what is the mass (grams) of the atomic surface? c. If the atomic surface were changed to ruthenium atoms and the same surface mass as determined in part \(\mathrm{b}\) is used, what number of ruthenium atoms is needed to construct the surface?

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