Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Balance the following equations and indicate whether they are combination, decomposition, or combustion reactions: $$ \begin{array}{l}{\text { (a) } \mathrm{PbCO}_{3}(s) \longrightarrow \mathrm{PbO}(s)+\mathrm{CO}_{2}(g)} \\ {\text { (b) } \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)} \\ {\text { (c) } \mathrm{Mg}(s)+\mathrm{N}_{2}(g) \longrightarrow \mathrm{Mg}_{3} \mathrm{N}_{2}(s)} \\ {\text { (d) } \mathrm{C}_{7} \mathrm{H}_{8} \mathrm{O}_{2}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)} \\ \text { (e) } \mathrm{Al}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{AlCl}_{3}(s) \end{array} $$

Short Answer

Expert verified
(a) \(PbCO_3(s) \longrightarrow PbO(s) + CO_2(g)\) (Decomposition Reaction) (b) \(C_2H_4(g) + 3O_2(g) \longrightarrow 2CO_2(g) + 2H_2O(g)\) (Combustion Reaction) (c) \(6Mg(s) + N_2(g) \longrightarrow 2Mg_3N_2(s)\) (Combination Reaction) (d) \(C_7H_8O_2(l) + 9O_2(g) \longrightarrow 7CO_2(g) + 4H_2O(g)\) (Combustion Reaction) (e) \(2Al(s) + 3Cl_2(g) \longrightarrow 2AlCl_3(s)\) (Combination Reaction)

Step by step solution

01

(a) Balancing the equation and determining the reaction type

: Given equation: \(PbCO_3(s) \longrightarrow PbO(s) + CO_2(g)\) This equation seems to be properly balanced, so we can directly analyze its reaction type. As a single reactant breaks down into multiple products, it is a decomposition reaction. Balanced equation and type: \(PbCO_3(s) \longrightarrow PbO(s) + CO_2(g)\) (Decomposition Reaction)
02

(b) Balancing the equation and determining the reaction type

: Given equation: \(C_2H_4(g) + O_2(g) \longrightarrow CO_2(g) + H_2O(g)\) To balance this equation, we need to start by adjusting the coefficients of the compounds. Step 1: Balance the carbon atoms. \(C_2H_4(g) + O_2(g) \longrightarrow 2CO_2(g) + H_2O(g)\) Step 2: Balance the hydrogen atoms. \(C_2H_4(g) + O_2(g) \longrightarrow 2CO_2(g) + 2H_2O(g)\) Step 3: Balance the oxygen atoms. \(C_2H_4(g) + 3O_2(g) \longrightarrow 2CO_2(g) + 2H_2O(g)\) The equation is now balanced. As we can observe reactants combining to produce one or more complex products, so this is a combustion reaction. Balanced equation and type: \(C_2H_4(g) + 3O_2(g) \longrightarrow 2CO_2(g) + 2H_2O(g)\) (Combustion Reaction)
03

(c) Balancing the equation and determining the reaction type

: Given equation: \(Mg(s) + N_2(g) \longrightarrow Mg_3N_2(s)\) To balance this equation, we need to adjust the coefficients of the compounds. Step 1: Balance the magnesium atoms. \[ 3Mg(s) + N_2(g) \longrightarrow Mg_3N_2(s) \] Step 2: Balance the nitrogen atoms. \[ 3Mg(s) + 1/2N_2(g) \longrightarrow Mg_3N_2(s) \] To avoid fractions, we can multiply the entire equation by 2, \[ 6Mg(s) + N_2(g) \longrightarrow 2Mg_3N_2(s) \] The equation is now balanced. As two simpler substances combine to form a single, more complex compound, this is a combination reaction. Balanced equation and type: \(6Mg(s) + N_2(g) \longrightarrow 2Mg_3N_2(s)\) (Combination Reaction)
04

(d) Balancing the equation and determining the reaction type

: Given equation: \(C_7H_8O_2(l) + O_2(g) \longrightarrow CO_2(g) + H_2O(g)\) To balance this equation, we need to adjust the coefficients of the compounds. Step 1: Balance the carbon atoms. \(C_7H_8O_2(l) + O_2(g) \longrightarrow 7CO_2(g) + H_2O(g)\) Step 2: Balance the hydrogen atoms. \(C_7H_8O_2(l) + O_2(g) \longrightarrow 7CO_2(g) + 4H_2O(g)\) Step 3: Balance the oxygen atoms. \(C_7H_8O_2(l) + 9O_2(g) \longrightarrow 7CO_2(g) + 4H_2O(g)\) The equation is now balanced. As we can see reactants combining to form one or more complex products, this is a combustion reaction. Balanced equation and type: \(C_7H_8O_2(l) + 9O_2(g) \longrightarrow 7CO_2(g) + 4H_2O(g)\) (Combustion Reaction)
05

(e) Balancing the equation and determining the reaction type

: Given equation: \(Al(s) + Cl_2(g) \longrightarrow AlCl_3(s)\) To balance this equation, we need to adjust the coefficients of the compounds. Step 1: Balance the aluminum atoms. \[ 2Al(s) + Cl_2(g) \longrightarrow 2AlCl_3(s) \] Step 2: Balance the chlorine atoms. \[ 2Al(s) + 3Cl_2(g) \longrightarrow 2AlCl_3(s) \] The equation is now balanced. As two simpler substances combine to form a single, more complex compound, this is a combination reaction. Balanced equation and type: \(2Al(s) + 3Cl_2(g) \longrightarrow 2AlCl_3(s)\) (Combination Reaction)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combustion Reactions
Combustion reactions are an exciting and frequently encountered type of chemical reaction where a substance combines with oxygen, releasing energy in the form of light or heat. These reactions are typically exothermic, meaning they give off energy rather than consume it. A common example that you might be familiar with is the burning of a candle. The wax of the candle reacts with oxygen in the air, producing carbon dioxide and water vapor as products.
Some key characteristics of combustion reactions include:
  • They involve oxygen as a reactant.
  • They release heat and sometimes light energy.
  • They often result in the formation of carbon dioxide and water.
In the context of balancing equations, it is crucial to ensure that the number of atoms for each element is equal on both sides. This balancing act can occasionally be tricky, especially with hydrocarbons, where both carbon and hydrogen need careful consideration. For instance, the combustion of ethene, which is given by the equation \(C_2H_4(g) + 3O_2(g) \to 2CO_2(g) + 2H_2O(g)\), showcases a typical combustion process where you have to ensure the equation is balanced with an equal number of carbon, hydrogen, and oxygen atoms on each side.
Decomposition Reactions
When we talk about decomposition reactions, we refer to a single chemical compound breaking down into two or more simpler substances. This kind of reaction is crucial in various biological and industrial processes. An example of a decomposition reaction is the breakdown of water into hydrogen and oxygen gases.Decomposition reactions commonly require an external source of energy, such as heat, light, or electricity. Here are some points to remember:
  • These reactions often require energy input to occur.
  • They break a compound into multiple products.
  • These reactions are critical in recycling materials within the environment.
In the provided exercise, the reaction \(PbCO_3(s) \rightarrow PbO(s) + CO_2(g)\) is a simple example of a decomposition reaction. The compound lead carbonate (\(PbCO_3\)) splits into lead oxide (\(PbO\)) and carbon dioxide (\(CO_2\)), showcasing how one compound converts into simpler constituents under appropriate conditions.
Combination Reactions
Combination reactions are the opposite of decomposition reactions. Here, two or more substances combine to form a single, more complex product. This type of reaction is quite common and forms the basis for creating many compounds, both naturally and synthetically.Key features of combination reactions include:
  • Multiple reactants form a single product.
  • The pure substances often react in such a way that they form entirely new compounds.
  • These reactions are usually straightforward to balance since fewer products and their compositions are involved.
A good example given in the exercise is \(6Mg(s) + N_2(g) \rightarrow 2Mg_3N_2(s)\). Here, magnesium and nitrogen gases react to form magnesium nitride, illustrating how two elements combine into a single chemical compound. This type of reaction is also commonly seen in the formation of salts, metals reacting with non-metals to create stable compounds.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The fizz produced when an Alka-Seltzer tablet is dissolved in water is due to the reaction between sodium bicarbonate \(\left(\mathrm{NaHCO}_{3}\right)\) and citric acid \(\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right) :\) $$ \begin{aligned} 3 \mathrm{NaHCO}_{3}(a q)+\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) & \longrightarrow \\ & 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) \end{aligned} $$ In a certain experiment 1.00 g of sodium bicarbonate and 1.00 g of citric acid are allowed to react. (a) Which is the limiting reactant? (b) How many grams of carbon dioxide form? (c) How many grams of the excess reactant remain after the limiting reactant is completely consumed?

Balance the following equations: $$ \begin{array}{l}{\text { (a) } \mathrm{Al}_{4} \mathrm{C}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{CH}_{4}(g)} \\ {\text { (b) } \mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{2}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)} \\ {\text { (c) } \mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)} \\ {\text { (d) } \mathrm{Mg}_{3} \mathrm{N}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{MgSO}_{4}(a q)+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(a q)}\end{array} $$

Balance the following equations: $$ \begin{array}{l}{\text { (a) } \mathrm{Li}(s)+\mathrm{N}_{2}(g) \longrightarrow \mathrm{Li}_{3} \mathrm{N}(s)} \\ {\text { (b) } \mathrm{TiCl}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{TiO}_{2}(s)+\mathrm{HCl}(a q)} \\ {\text { (c) } \mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)} \\ {\text { (d) } \mathrm{AlCl}_{3}(s)+\mathrm{Ca}_{3} \mathrm{N}_{2}(s) \longrightarrow \mathrm{AlN}(s)+\mathrm{CaCl}_{2}(s)}\end{array} $$

Determine the empirical formulas of the compounds with the following compositions by mass: $$ \begin{array}{l}{\text { (a) } 10.4 \% \mathrm{C}, 27.8 \% \mathrm{S}, \text { and } 61.7 \% \mathrm{Cl}} \\ {\text { (b) } 21.7 \% \mathrm{C}, 9.6 \% \mathrm{O}, \text { and } 68.7 \% \mathrm{F}} \\ {\text { (c) } 32.79 \% \mathrm{Na}, 13.02 \% \mathrm{Al}, \text { and the remainder } \mathrm{F}}\end{array} $$

Propenoic acid, \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{2},\) is a reactive organic liquid that is used in the manufacturing of plastics, coatings, and adhesives. An unlabeled container is thought to contain this liquid. A 0.275 -g sample of the liquid is combusted to produce 0.102 gof water and 0.374 g carbon dioxide. Is the unknown liquid propenoic acid? Support your reasoning with calculations.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free