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Chapter 5: Problem 97

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, catalyzed by manganese(IV) oxide, is: \( 2 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{O}_{2} + 2 \mathrm{H}_{2} \mathrm{O} \)

Step by step solution

01

Write the unbalanced equation

To begin, let's write down the unbalanced equation for the reaction, with reactants on the left and products on the right: H鈧侽鈧� 鈫� O鈧� + H鈧侽 This equation represents the decomposition of hydrogen peroxide into oxygen gas and water.
02

Checking for catalyst

Since manganese(IV) oxide is a catalyst, it does not take part in the reaction, and thus it will not be included in the balanced chemical equation.
03

Balancing the equation

Now, let's balance the chemical equation. We'll start by counting the atoms of each element on the reactant and product sides: Reactants: 2 H atoms 2 O atoms Products: 2 H atoms 2 + 1 = 3 O atoms The hydrogen atoms are already balanced, but there is one extra oxygen atom in the products. To balance the equation, we can add coefficients before the chemical species to equalize the number of atoms for each element: 2 H鈧侽鈧� 鈫� O鈧� + 2 H鈧侽 Now, we have: Reactants: 4 H atoms 4 O atoms Products: 4 H atoms 2 + 2 = 4 O atoms The equation is now balanced.
04

Write the balanced equation

The balanced equation for the decomposition reaction of hydrogen peroxide is: 2 H鈧侽鈧� 鈫� O鈧� + 2 H鈧侽

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Most popular questions from this chapter

Elixirs such as Alka-Seltzer use the reaction of sodium bicarbonate with citric acid in aqueous solution to produce a fizz: $$\begin{aligned}3 \mathrm{NaHCO}_{3}(a q)+\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}(a q) & \longrightarrow \\\3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) &+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q)\end{aligned}$$ a. What mass of \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\) should be used for every \(1.0 \times 10^{2} \mathrm{mg} \mathrm{NaHCO}_{3} ?\) b. What mass of \(\mathrm{CO}_{2}(g)\) could be produced from such a mixture?

Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$$ a. What is the maximum mass of ammonia that can be produced from a mixture of \(1.00 \times 10^{3} \mathrm{g} \mathrm{N}_{2}\) and \(5.00 \times 10^{2} \mathrm{g} \mathrm{H}_{2} ?\) b. What mass of which starting material would remain unreacted?

Commercial brass, an alloy of Zn and Cu, reacts with hydrochloric acid as follows: $$\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ (Cu does not react with HCl.) When 0.5065 g of a certain brass alloy is reacted with excess \(\mathrm{HCl}, 0.0985 \mathrm{g}\) \(\mathrm{ZnCl}_{2}\) is eventually isolated. a. What is the composition of the brass by mass? b. How could this result be checked without changing the above procedure?

A given sample of a xenon fluoride compound contains molecules of the type \(\mathrm{XeF}_{n},\) where \(n\) is some whole number. Given that \(9.03 \times 10^{20}\) molecules of \(\mathrm{XeF}_{n}\) weigh \(0.368 \mathrm{g},\) determine the value for \(n\) in the formula.

Express the composition of each of the following compounds as the mass percents of its elements. a. formaldehyde, \(\mathrm{CH}_{2} \mathrm{O}\) b. glucose, \(C_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) c. acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)
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