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Balance the following equations: (a) \(\mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})\) (b) \(\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HNO}_{3}(a q)\) (c) \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(l)+\mathrm{HCl}(g)\) (d) \(\mathrm{Al}_{4} \mathrm{C}_{3}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{CH}_{4}(g)\) (e) \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{2}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (f) \(\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q)\) \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (g) \(\mathrm{Mg}_{3} \mathrm{~N}_{2}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) \(\mathrm{MgSO}_{4}(a q)+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(a q)\)

Short Answer

Expert verified
Balanced equations: (a) \(CO(g) + O_2(g) \longrightarrow CO_2(g)\) (b) \(N_2O_5(g) + H_2O(l) \longrightarrow 2HNO_3(aq)\) (c) \(CH_4(g) + 2Cl_2(g) \longrightarrow CCl_4(l) + 4HCl(g)\) (d) \(Al_4C_3(s) + 12H_2O(l) \longrightarrow 4Al(OH)_3(s) + 3CH_4(g)\) (e) \(C_5H_{10}O_2(l) + 7.5O_2(g) \longrightarrow 5CO_2(g) + 5H_2O(g)\) (f) \(2Fe(OH)_3(s) + 3H_2SO_4(aq) \longrightarrow Fe_2(SO_4)_3(aq) + 6H_2O(l)\) (g) \(Mg_3N_2(s) + 6H_2SO_4(aq) \longrightarrow 3MgSO_4(aq) + 2(NH_4)_2SO_4(aq)\)

Step by step solution

01

List the atoms on both sides of the equation

Identify the atoms present in both the reactants and products. In this case, we have carbon (C) and oxygen (O).
02

Count the atoms for each element on both sides

Reactants: C - 1, O - 2 Products: C - 1, O - 2
03

Balance the equation

The equation is already balanced: \(CO(g) + O_2(g) \longrightarrow CO_2(g)\) (b) Balance the equation: \(N_2O_5(g) + H_2O(l) \longrightarrow HNO_3(aq)\)
04

List the atoms on both sides of the equation

Identify the atoms present in both the reactants and products. In this case, we have nitrogen (N), oxygen (O), and hydrogen (H).
05

Count the atoms for each element on both sides

Reactants: N - 2, O - 5, H - 2 Products: N - 1, O - 4, H - 1
06

Balance the equation

To balance the equation, we need to add a coefficient of 2 in front of \(HNO_3\). The balanced equation will be: \(N_2O_5(g) + H_2O(l) \longrightarrow 2HNO_3(aq)\) (c) Balance the equation: \(CH_4(g) + Cl_2(g) \longrightarrow CCl_4(l) + HCl(g)\)
07

List the atoms on both sides of the equation

Identify the atoms present in both the reactants and products. In this case, we have carbon (C), hydrogen (H), and chlorine (Cl).
08

Count the atoms for each element on both sides

Reactants: C - 1, H - 4, Cl - 2 Products: C - 1, H - 1, Cl - 5
09

Balance the equation

To balance the equation, we need to add coefficients of 4 in front of \(HCl\), and 2 in front of \(Cl_2\). The balanced equation will be: \(CH_4(g) + 2Cl_2(g) \longrightarrow CCl_4(l) + 4HCl(g)\) We will follow the same steps for the remaining equations to balance them. (d) \(Al_4C_3(s) + H_2O(l) \longrightarrow Al(OH)_3(s) + CH_4(g)\) Balanced equation: \(Al_4C_3(s) + 12H_2O(l) \longrightarrow 4Al(OH)_3(s) + 3CH_4(g)\) (e) \(C_5H_{10}O_2(l) + O_2(g) \longrightarrow CO_2(g) + H_2O(g)\) Balanced equation: \(C_5H_{10}O_2(l) + 7.5O_2(g) \longrightarrow 5CO_2(g) + 5H_2O(g)\) (f) \(Fe(OH)_3(s) + H_2SO_4(aq) \longrightarrow Fe_2(SO_4)_3(aq) + H_2O(l)\) Balanced equation: \(2Fe(OH)_3(s) + 3H_2SO_4(aq) \longrightarrow Fe_2(SO_4)_3(aq) + 6H_2O(l)\) (g) \(Mg_3N_2(s) + H_2SO_4(aq) \longrightarrow MgSO_4(aq) + (NH_4)_2SO_4(aq)\) Balanced equation: \(Mg_3N_2(s) + 6H_2SO_4(aq) \longrightarrow 3MgSO_4(aq) + 2(NH_4)_2SO_4(aq)\)

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Key Concepts

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

Stoichiometry
Stoichiometry is like the recipe of chemistry. It's about measuring the amounts of each reactant and product in a chemical reaction. When we talk about stoichiometry, we are essentially discussing the exact ratio of molecules or atoms needed to react with each other in a chemical reaction. This is pivotal in understanding chemical reactions because it helps us to predict how much of a substance will be produced in a reaction given a certain amount of reactants.
For instance, in the given equation - \(\mathrm{CH}_{4}(g) + 2\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(l) + 4\mathrm{HCl}(g)\),- stoichiometry ensures that the correct amount of chloride reacts with methane. Every single molecule of \(\mathrm{CH}_{4}\) reacts precisely with exactly two \(\mathrm{Cl}_{2}\) molecules to form the products without any left over, respecting the stoichiometric coefficients in the equation. This balance maintains the law of conservation of mass, which we'll explore next.
Law of Conservation of Mass
The law of conservation of mass is a fundamental concept in chemistry. It states that in any chemical reaction, mass is neither created nor destroyed. This principle is the reason why we balance chemical equations. When a chemical equation is balanced, it means that the mass of the reactants is equal to the mass of the products.
For example, when you balance the equation - \(\mathrm{N}_{2}\mathrm{O}_{5}(g) + \mathrm{H}_{2}\mathrm{O}(l) \longrightarrow 2\mathrm{HNO}_{3}(aq)\),- the mass of the nitrogen, oxygen, and hydrogen atoms on the left side of the equation is the same as on the right side. This balanced state reflects the conservation of mass. Each atom that starts in the reactants ends up in the products, ensuring that no mass is lost during the process. Understanding this law is crucial for making accurate predictions and measurements in chemistry.
Chemical Reactions
Chemical reactions are processes where substances, known as reactants, are transformed into new substances, called products. These transformations occur through the rearrangement of atoms. Each reaction involves breaking and forming bonds, which can be observed through changes in properties, such as color or temperature.
The reaction - \(\mathrm{Al}_{4}\mathrm{C}_{3}(s) + 12\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow 4\mathrm{Al}(\mathrm{OH})_{3}(s) + 3\mathrm{CH}_{4}(g)\),- is a good example. Here, atoms from aluminum carbide and water react to form aluminum hydroxide and methane gas. This process involves breaking bonds in the reactants and forming new bonds in the products.
Chemical reactions can be classified into types such as combination, decomposition, substitution, and double displacement. Each type has its unique patterns for rearranging atoms. Recognizing these patterns is useful in predicting the products of an unknown chemical reaction.
Coefficient Adjustment
Coefficient adjustment is a crucial step in balancing chemical equations. It involves changing the numbers in front of formulas in the equation to ensure that the number of atoms for each element is the same on both sides of the equation. This adjustment does not alter the identity of the substances, but rather, ensures the equation respects the stoichiometry and the law of conservation of mass.
Consider the equation - \(\mathrm{Fe}(\mathrm{OH})_{3}(s) + \mathrm{H}_{2}\mathrm{SO}_{4}(aq) \longrightarrow \mathrm{Fe}_{2}(\mathrm{SO}_{4})_{3}(aq) + \mathrm{H}_{2}\mathrm{O}(l)\).- By adjusting the coefficients, we apply - "2" in front of \(\mathrm{Fe}(\mathrm{OH})_{3}\) and "3" in front of \(\mathrm{H}_{2}\mathrm{SO}_{4}\),- resulting in the balanced form - \(2\mathrm{Fe}(\mathrm{OH})_{3}(s) + 3\mathrm{H}_{2}\mathrm{SO}_{4}(aq) \longrightarrow \mathrm{Fe}_{2}(\mathrm{SO}_{4})_{3}(aq) + 6\mathrm{H}_{2}\mathrm{O}(l)\).- This process ensures the reaction follows chemical laws, especially the conservation of mass. Coefficient adjustment makes chemical equations not just mathematically equal but chemically accurate.

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Most popular questions from this chapter

When benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) reacts with bromine \(\left(\mathrm{Br}_{2}\right)\), bromobenzene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}\right)\) is obtained: $$ \mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{Br}_{2} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}+\mathrm{HBr} $$ (a) What is the theoretical yield of bromobenzene in this reaction when \(30.0 \mathrm{~g}\) of benzene reacts with \(65.0 \mathrm{~g}\) of bromine? (b) If the actual yield of bromobenzene was \(42.3 \mathrm{~g}\), what was the percentage yield?

Determine the empirical and molecular formulas of each of the following substances: (a) Styrene, a compound substance used to make Styrofoam \(^{B}\) cups and insulation, contains \(92.3 \% \mathrm{C}\) and \(7.7 \% \mathrm{H}\) by mass and has a molar mass of \(104 \mathrm{~g} / \mathrm{mol}\). (b) Caffeine, a stimulant found in coffee, contains \(49.5 \%\) C, \(5.15 \% \mathrm{H}, 28.9 \% \mathrm{~N}\), and \(16.5 \% \mathrm{O}\) by mass and has a molar mass of \(195 \mathrm{~g} / \mathrm{mol}\). (c) Monosodium glutamate (MSG), a flavor enhancer in certain foods, contains \(35.51 \%\) C, \(4.77 \% \mathrm{H}, 37.85 \%\) O, \(8.29 \% \mathrm{~N}\), and \(13.60 \%\) Na, and has a molar mass of \(169 \mathrm{~g} / \mathrm{mol}\).

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) together with other substances. Reaction of the ore with CO produces iron metal: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$ (a) Balance this equation. (b) Calculate the number of grams of CO that can react with \(0.150 \mathrm{~kg}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\). (c) Calculate the number of grams of Fe and the number of grams of \(\mathrm{CO}_{2}\) formed when \(0.150 \mathrm{~kg}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) reacts. (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

When a mixture of \(10.0 \mathrm{~g}\) of acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and \(10.0 \mathrm{~g}\) of oxygen \(\left(\mathrm{O}_{2}\right)\) is ignited, the resultant combustion reaction produces \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} .\) (a) Write the balanced chemical equation for this reaction. (b) Which is the limiting reactant? (c) How many grams of \(\mathrm{C}_{2} \mathrm{H}_{2}, \mathrm{O}_{2}, \mathrm{CO}_{2}\), and \(\mathrm{H}_{2} \mathrm{O}\) are present after the reaction is complete?

Aluminum hydroxide reacts with sulfuric acid as follows: $$ 2 \mathrm{Al}(\mathrm{OH})_{3}(s)+3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ Which reagent is the limiting reactant when \(0.500 \mathrm{~mol}\) \(\mathrm{Al}(\mathrm{OH})_{3}\) and \(0.500 \mathrm{~mol} \mathrm{H}_{2} \mathrm{SO}_{4}\) are allowed to react? How many moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) can form under these conditions? How many moles of the excess reactant remain after the completion of the reaction?

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