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Chapter 8: Problem 28

Balance each of the following: $$\begin{array}{l}{\text { a. Fe }+\mathrm{O}_{2} \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}} \\ {\text { b. } \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}} \\ {\text { c. } \mathrm{C}_{8} \mathrm{H}_{18}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}} \\ {\text { d. } \mathrm{Al}+\mathrm{F}_{2} \rightarrow \mathrm{AlF}_{3}}\end{array} $$

Short Answer

Expert verified
a. \(2Fe+3O_{2} \rightarrow 2Fe_{2}O_{3}\), b. \(2H_{2}O_{2} \rightarrow 2H_{2}O+O_{2}\), c. \(2C_{8}H_{18}+25O_{2} \rightarrow 16CO_{2}+18H_{2}O\), d. \(2Al+3F_{2} \rightarrow 2AlF_{3}\)

Step by step solution

01

Balancing Iron and Oxygen

Looking at formula a. \(Fe+O_{2} \rightarrow Fe_{2}O_{3}\) - start by balancing iron (Fe) since there's only one type on each side. It is balanced by having 2 Fe on the right side, so add a coefficient 2 to the left side for Fe. Now there's 6 O on the right side (since \(Fe_{2}O_{3}\) has 3*2 O), add a 3 to \(O_{2}\) on the left side to balance it. The balanced equation: \(2Fe+3O_{2} \rightarrow 2Fe_{2}O_{3}\)
02

Balancing Hydrogen Peroxide to Water and Oxygen

For formula b. \(H_{2}O_{2} \rightarrow H_{2}O+O_{2}\) - start with Hydrogen (H). Adding a 2 before H2O balances it. Oxygen atoms can be balanced by simply placing a 2 before O2. The balanced equation is: \(2H_{2}O_{2} \rightarrow 2H_{2}O+O_{2}\)
03

Balancing Octane and Oxygen to Carbon Dioxide and Water

For formula c. \(C_{8}H_{18}+O_{2} \rightarrow CO_{2}+H_{2}O\) - The most complex of these. Always start balancing with the element that appears the least, so start by balancing C. For this, place an 8 in front of CO2 on the right. To balance H, place a 9 in front of \(H_{2}O\). Finally to balance O, count the total oxygen atoms in the right side which will be 25. In order to make the number of oxygen atoms equal on both sides, the number of \(O_{2}\) molecules must be 12.5 on left. But a chemical equation should have whole numbers, to achieve this, multiply entire equation by 2. The balanced equation is: \(2C_{8}H_{18}+25O_{2} \rightarrow 16CO_{2}+18H_{2}O\)
04

Balancing Aluminium and Fluorine

For formula d. \(Al+F_{2} \rightarrow AlF_{3}\) - looking at the right side, there are 3 F atoms, so we add a 3 before \(F_{2}\) on the left side. Then add a 2 before Al on the left and \(AlF_{3}\) on the right to balance aluminium. The balanced equation is: \(2Al+3F_{2} \rightarrow 2AlF_{3}\)

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Most popular questions from this chapter

The white paste that lifeguards rub on their nose to prevent sunburn contains zinc oxide, \(\mathrm{ZnO}(s),\) as an active ingredient. Zinc oxide is produced by burning zinc sulfide. $$ 2 \mathrm{ZnS}(s)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{ZnO}(s)+2 \mathrm{SO}_{2}(g) $$ $$\begin{array}{l}{\text { a. What is the coefficient for sulfur dioxide? }} \\\ {\text { b. What is the subscript for oxygen gas? }} \\ {\text { c. How many atoms of oxygen react? }} \\ {\text { d. How many atoms of oxygen react? }} \\ {\text { the total number of sulfur dioxide }} \\ {\text { molecules? }}\end{array}$$

Balance each of the following: $$\begin{array}{l}{\text { a. } \mathrm{Zn}+\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} \rightarrow \mathrm{Pb}+\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}} \\ {\text { b. } \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+\mathrm{NaOH} \rightarrow \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+\mathrm{H}_{2} \mathrm{O}} \\\ {\text { c. } \mathrm{Al}+\mathrm{CuSO}_{4} \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{Cu}}\end{array}$$

The following equations are incorrect in some way. Identify and correct each error, and then balance each equation. $$\begin{array}{l}{\text { a. } \mathrm{Li}+\mathrm{O}_{2} \rightarrow \mathrm{LiO}_{2}} \\ {\text { b. } \mathrm{MgCO}_{3} \rightarrow \mathrm{Mg}+\mathrm{C}+3 \mathrm{O}_{2}} \\ {\text { c. } \mathrm{NaI}+\mathrm{Cl}_{2} \rightarrow \mathrm{NaCl}+\mathrm{I}} \\ {\text { d. } \mathrm{AgNO}_{3}+\mathrm{CaCl}_{2} \rightarrow \mathrm{Ca}\left(\mathrm{NO}_{3}\right)+\mathrm{AgCl}_{2}} \\ {\text { e. } 3 \mathrm{Mg}+2 \mathrm{FeBr}_{3} \rightarrow \mathrm{Fe}_{2} \mathrm{Mg}_{3}+3 \mathrm{Br}_{2}}\end{array}$$

Write an unbalanced chemical equation for each of the following. $$\begin{array}{l}{\text { a. Aluminum reacts with oxygen to produce }} \\\ {\text { aluminum oxide. }} \\ {\text { b. Phosphoric acid, } \mathrm{H}_{3} \mathrm{PO}_{4} \text { , is produced }} \\ {\text { through the reaction between tetraphosphorus}} \\ {\text { decoxide and water. }}\end{array}$$

Write the total and net ionic equations for the reaction in which the antacid Al(OH) \(_{3}\) neutralizes the stomach acid HCl. Identify the type of reaction. $$\begin{array}{l}{\text { a. Identify the spectator ions in this reaction. }} \\\ {\text { b. What would be the advantages of using }} \\ {\text { Al(OH) }_{3} \text { as an antacid rather than }} \\ {\text { NaHCO_ }_{3} \text { , which undergoes the following }} \\ {\text { reaction with stomach acid? }}\end{array}$$ $$\begin{array}{c}{\mathrm{NaHCO}_{3}(a q)+\mathrm{HCl}(a q) \rightarrow}{\mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)}\end{array}$$
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