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: (a) \(\mathrm{Li}(s)+\mathrm{N}_{2}(g) \longrightarrow \mathrm{Li}_{3} \mathrm{~N}(s)\) (b) \(\mathrm{La}_{2} \mathrm{O}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{La}(\mathrm{OH})_{3}(a q)\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{Ca}_{3} \mathrm{P}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{PH}_{3}(g)\) (e) \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow\) \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) (f) \(\mathrm{AgNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(\mathrm{~s})+\mathrm{NaNO}_{3}(a q)\) (g) \(\mathrm{CH}_{3} \mathrm{NH}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow\) \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)\)

Short Answer

Expert verified
The balanced equations are: (a) \(6\mathrm{Li}(s) + \mathrm{N}_{2}(g) \longrightarrow 2\mathrm{Li}_{3}\mathrm{N}(s)\) (b) \(\mathrm{La}_{2}\mathrm{O}_{3}(s) + 6\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow 2\mathrm{La}(\mathrm{OH})_{3}(a q)\) (c) \(\mathrm{NH}_{4}\mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2}(g)+\frac{1}{2}\mathrm{O}_{2}(g)+2\mathrm{H}_{2}\mathrm{O}(g)\) (d) \(\mathrm{Ca}_{3}\mathrm{P}_{2}(s) + 6\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow 3\mathrm{Ca}(\mathrm{OH})_{2}(a q)+2\mathrm{PH}_{3}(g)\) (e) \(3\mathrm{Ca}(\mathrm{OH})_{2}(a q) + 2\mathrm{H}_{3}\mathrm{PO}_{4}(a q) \longrightarrow 1\mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2}(s) + 6\mathrm{H}_{2}\mathrm{O}(l)\) (f) \(2\mathrm{AgNO}_{3}(a q)+\mathrm{Na}_{2}\mathrm{SO}_{4}(a q) \longrightarrow 1\mathrm{Ag}_{2}\mathrm{SO}_{4}(\mathrm{s})+2\mathrm{NaNO}_{3}(a q)\) (g) \(\mathrm{CH}_{3}\mathrm{NH}_{2}(g) + \frac{3}{2}\mathrm{O}_{2}(g) \longrightarrow 1\mathrm{CO}_{2}(g)+2\mathrm{H}_{2}\mathrm{O}(g)+\frac{1}{2}\mathrm{N}_{2}(g)\)

Step by step solution

01

Identifying the Elements

Here, we have Lithium \(\mathrm{(Li)}\) and Nitrogen \(\mathrm{(N)}\).
02

Counting Atoms

There are 1 \(\mathrm{Li}\) and 2 \(\mathrm{N}\) atoms on the left-hand side, and 3 \(\mathrm{Li}\) and 1 \(\mathrm{N}\) atom on the right-hand side.
03

Adjusting Coefficients

To balance the equation, we'll adjust the coefficients as follows: \(6\mathrm{Li}(s) + \mathrm{N}_{2}(g) \longrightarrow 2\mathrm{Li}_{3}\mathrm{N}(s)\) (b) Balancing \(\mathrm{La}_{2}\mathrm{O}_{3}(s)+\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{La}(\mathrm{OH})_{3}(a q)\)
04

Identifying the Elements

Here, we have Lanthanum \(\mathrm{(La)}\), Oxygen \(\mathrm{(O)}\), and Hydrogen \(\mathrm{(H)}\).
05

Counting Atoms

There are 2 \(\mathrm{La}\), 3 \(\mathrm{O}\), and 2 \(\mathrm{H}\) atoms on the left-hand side, and 1 \(\mathrm{La}\), 4 \(\mathrm{O}\), and 3 \(\mathrm{H}\) atoms on the right-hand side.
06

Adjusting Coefficients

To balance the equation, we'll adjust the coefficients as follows: \(\mathrm{La}_{2}\mathrm{O}_{3}(s) + 6\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow 2\mathrm{La}(\mathrm{OH})_{3}(a q)\) (c) Balancing \(\mathrm{NH}_{4}\mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{H}_{2}\mathrm{O}(g)\)
07

Identifying the Elements

Here, we have Nitrogen \((N)\), Oxygen \((O)\), and Hydrogen \((H)\).
08

Counting Atoms

There are 2 \(\mathrm{N}\), 4 \(\mathrm{O}\), and 4 \(\mathrm{H}\) atoms on the left-hand side, and 2 \(\mathrm{N}\), 2 \(\mathrm{O}\), and 2 \(\mathrm{H}\) atoms on the right-hand side.
09

Adjusting Coefficients

To balance the equation, we'll adjust the coefficients as follows: \(\mathrm{NH}_{4}\mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2}(g)+\frac{1}{2}\mathrm{O}_{2}(g)+2\mathrm{H}_{2}\mathrm{O}(g)\) (d) Balancing \(\mathrm{Ca}_{3}\mathrm{P}_{2}(s)+\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{PH}_{3}(g)\)
10

Identifying the Elements

Here, we have Calcium \((Ca)\), Phosphorus \((P)\), Oxygen \((O)\), and Hydrogen \((H)\).
11

Counting Atoms

There are 3 \(\mathrm{Ca}\), 2 \(\mathrm{P}\), 2 \(\mathrm{H}\), and 1 \(\mathrm{O}\) atoms on the left-hand side, and 1 \(\mathrm{Ca}\), 1 \(\mathrm{P}\), 2 \(\mathrm{O}\), and 5 \(\mathrm{H}\) atoms on the right-hand side.
12

Adjusting Coefficients

To balance the equation, we'll adjust the coefficients as follows: \(\mathrm{Ca}_{3}\mathrm{P}_{2}(s) + 6\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow 3\mathrm{Ca}(\mathrm{OH})_{2}(a q)+2\mathrm{PH}_{3}(g)\) (e) Balancing \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{H}_{3}\mathrm{PO}_{4}(a q) \longrightarrow \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2}(s)+\mathrm{H}_{2}\mathrm{O}(l)\)
13

Identifying the Elements

Here, we have Calcium \((Ca)\), Oxygen \((O)\), Hydrogen \((H)\), and Phosphorus \((P)\).
14

Counting Atoms

There are 1 \(\mathrm{Ca}\), 2 \(\mathrm{O}\), 5 \(\mathrm{H}\), and 1 \(\mathrm{P}\) atoms on the left-hand side, and 3 \(\mathrm{Ca}\), 8 \(\mathrm{O}\), 2 \(\mathrm{H}\), and 2 \(\mathrm{P}\) atoms on the right-hand side.
15

Adjusting Coefficients

To balance the equation, we'll adjust the coefficients as follows: \(3\mathrm{Ca}(\mathrm{OH})_{2}(a q) + 2\mathrm{H}_{3}\mathrm{PO}_{4}(a q) \longrightarrow 1\mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2}(s) + 6\mathrm{H}_{2}\mathrm{O}(l)\) (f) Balancing \(\mathrm{AgNO}_{3}(a q)+\mathrm{Na}_{2}\mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Ag}_{2}\mathrm{SO}_{4}(\mathrm{s})+\mathrm{NaNO}_{3}(a q)\)
16

Identifying the Elements

Here, we have Silver \((Ag)\), Sodium \((Na)\), Nitrogen \((N)\), and Oxygen \((O)\).
17

Counting Atoms

There are 1 \(\mathrm{Ag}\), 1 \(\mathrm{Na}\), 2 \(\mathrm{O}\), and 1 \(\mathrm{N}\) atoms on the left-hand side, and 2 \(\mathrm{Ag}\), 1 \(\mathrm{Na}\), 5 \(\mathrm{O}\), and 1 \(\mathrm{N}\) atoms on the right-hand side.
18

Adjusting Coefficients

To balance the equation, we'll adjust the coefficients as follows: \(2\mathrm{AgNO}_{3}(a q)+\mathrm{Na}_{2}\mathrm{SO}_{4}(a q) \longrightarrow 1\mathrm{Ag}_{2}\mathrm{SO}_{4}(\mathrm{s})+2\mathrm{NaNO}_{3}(a q)\) (g) Balancing \(\mathrm{CH}_{3}\mathrm{NH}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2}\mathrm{O}(g)+\mathrm{N}_{2}(g)\)
19

Identifying the Elements

Here, we have Carbon \((C)\), Hydrogen \((H)\), Nitrogen \((N)\), and Oxygen \((O)\).
20

Counting Atoms

There are 1 \(\mathrm{C}\), 5 \(\mathrm{H}\), 1 \(\mathrm{N}\), and 2 \(\mathrm{O}\) atoms on the left-hand side, and 1 \(\mathrm{C}\), 2 \(\mathrm{H}\), 2 \(\mathrm{N}\), and 3 \(\mathrm{O}\) atoms on the right-hand side.
21

Adjusting Coefficients

To balance the equation, we'll adjust the coefficients as follows: \(\mathrm{CH}_{3}\mathrm{NH}_{2}(g) + \frac{3}{2}\mathrm{O}_{2}(g) \longrightarrow 1\mathrm{CO}_{2}(g)+2\mathrm{H}_{2}\mathrm{O}(g)+\frac{1}{2}\mathrm{N}_{2}(g)\)

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.

Stoichiometry
Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is based on the law of conservation of mass, which states that in a chemical reaction, matter is neither created nor destroyed. This means that the total mass of the reactants must equal the total mass of the products.

Stoichiometry allows chemists to predict the amounts of substances consumed and produced in a reaction, enabling them to calculate yields and determine the optimal ratios of reactants. To perform stoichiometric calculations, it is necessary to have a balanced chemical equation. The coefficients in a balanced chemical equation indicate the ratios in which reactants and products react and are formed, respectively. These coefficients can also be viewed as the moles of each substance, connecting quantities to the balanced equation.

For example, in the exercise balancing \(\mathrm{Li}(s) + \mathrm{N}_{2}(g) \longrightarrow \mathrm{Li}_{3}\mathrm{N}(s)\), the balanced equation tells us that 6 moles of lithium react with 1 mole of nitrogen gas to form 2 moles of lithium nitride. In other exercises, adjusting coefficients may involve using fractions, as seen with \(\mathrm{NH}_{4}\mathrm{NO}_{3}(s)\), to respect the conservation of atoms within the reaction.
Chemical Reaction Balancing
Balancing chemical equations is a fundamental task in chemistry, as it ensures that the same number of each type of atom is present on both sides of the equation, adhering to the law of conservation of mass. The balancing act requires a methodical approach, typically starting with the most complex molecule and moving on to simpler substances.

To balance a chemical equation, one must adjust the coefficients in front of each chemical formula without altering the actual formulas. These coefficients represent the number of units (molecules, formula units, or moles) of each substance, and changing them alters the proportions in which substances react. The process often involves trial and error, as seen in the provided exercises, to find the smallest whole numbers that satisfy the balance of atoms.

As highlighted in the step-by-step solution, identifying elements and counting atoms in each is the starting point. After establishing the imbalances, adjusting coefficients methodically ensures each element has the same number of atoms on both sides of the reaction. In practical terms, developing a skill for balancing equations is essential for laboratory work, industrial processes, and environmental assessments where knowing exact proportions of chemicals is vital.
Chemical Formulation
Chemical formulation is the process of designing and creating chemical compounds with specific properties and compositions. In the context of balancing chemical equations, the term refers to the representation of compounds through their chemical formulas, which show the elements and the number of each type of atom present in the compound.

Chemical formulas are a shorthand way of describing the structure of molecules, and they play a crucial role in stoichiometry and the balancing of chemical reactions. Each element in the formula is represented by its chemical symbol, and subscripts are used to indicate the number of atoms of that element in each molecule or formula unit. For polyatomic ions and complexes, parentheses may be used to show the grouping of atoms.

For instance, in the balancing of \(\mathrm{Ca}_{3}\mathrm{P}_{2}(s) + \mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q) + \mathrm{PH}_{3}(g)\), the formulas \(\mathrm{Ca}_{3}\mathrm{P}_{2}\) and \(\mathrm{Ca}(\mathrm{OH})_{2}\) clearly indicate the composition of each compound. Students can use this information to identify necessary changes to coefficients in order to balance the equation. Understanding chemical formulation is critical to interpreting reaction mechanisms, predicting product formation, and effectively communicating chemical information.

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

Withoutdoing any detailed calculations (but using a periodic table to give atomic weights), rank the following samples in order of increasing number of atoms: \(3.0 \times 10^{23}\) molecules of \(\mathrm{H}_{2} \mathrm{O}_{2}, 2.0 \mathrm{~mol} \mathrm{CH}_{4}, 32 \mathrm{~g} \mathrm{O}_{2}\)

(a) What is the mass, in grams, of \(0.0714\) mol of iron(III) sulfate? (b) How many moles of ammonium ions are in \(8.776 \mathrm{~g}\) of ammonium carbonate? (c) What is the mass, in grams, of \(6.52 \times 10^{21}\) molecules of aspirin, \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4} ?\) (d) What is the molar mass of diazepam (Valium \(^{\otimes}\) ) if \(0.05570\) mol weighs \(15.86 \mathrm{~g}\) ?

Without doing any detailed calculations (but using a periodic table to give atomic weights), rank the following samples in order of increasing number of atoms: \(0.50 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}, 23 \mathrm{~g} \mathrm{Na}, 6.0 \times 10^{23} \mathrm{~N}_{2}\) molecules.

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?

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of \(\mathrm{NH}_{3}\) to \(\mathrm{NO}\) : $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ In a certain experiment, \(1.50 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) reacts with \(2.75 \mathrm{~g}\) of \(\mathrm{O}_{2}\) (a) Which is the limiting reactant? (b) How many grams of \(\mathrm{NO}\) and of \(\mathrm{H}_{2} \mathrm{O}\) form? (c) How many grams of the excess reactant remain after the limiting reactant is completely consumed? (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

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