Question

Balance the following equations: (a) \(\mathrm{L}_{\mathrm{L}(s)}+\mathrm{N}_{2}(g) \longrightarrow \mathrm{Li}_{3} \mathrm{~N}(s)\) (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) \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q)\) \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) (f) \(\mathrm{AgNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q) \overrightarrow{\mathrm{Ag}_{2} \mathrm{SO}_{4}(s)+\mathrm{NaNO}_{3}(a q)}\) (g) \(\mathrm{CH}_{3} \mathrm{NH}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \longrightarrow\) \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)\)

Short answer

The balanced equations are: (a) \(6 \mathrm{Li}_{\mathrm{L}(s)}+\mathrm{N}_{2}(g) \longrightarrow 2\mathrm{Li}_{3} \mathrm{~N}(s)\) (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) \(2\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2}(g)+3\mathrm{O}_{2}(g)+4\mathrm{H}_{2} \mathrm{O}(g)\) (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) \(3\mathrm{Ca}(\mathrm{OH})_{2}(a q)+2\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+6\mathrm{H}_{2}\mathrm{O}(l)\) (f) \(2\mathrm{AgNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Ag}_{2} \mathrm{SO}_{4}(s)+2\mathrm{NaNO}_{3}(a q)\) (g) \(\mathrm{CH}_{3} \mathrm{NH}_{2}(\mathrm{~g})+3\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)\)

Step-by-step solution

Identify the unbalanced atoms

The number of lithium atoms and nitrogen atoms in the reactants is not equal to the number of lithium atoms and nitrogen atoms in the products. We have one lithium atom and two nitrogen atoms in the reactants, but three lithium atoms and one nitrogen atom in the product.

Balance the equation

To balance the equation, we need to adjust the coefficients of the reactants and products: - Multiply the lithium atoms on the left side by 3 by adding a coefficient of 6 in front of \(\mathrm{Li_{L}(s)}\). - Multiply the nitrogen atoms on the right side by 2 by adding a coefficient of 2 in front of \(\mathrm{Li}_{3} \mathrm{~N}(s)\). The balanced equation is: \(6 \mathrm{Li_{L}(s)}+\mathrm{N}_{2}(g) \longrightarrow 2\mathrm{Li}_{3} \mathrm{~N}(s)\) (b) \(\mathrm{La}_{2} \mathrm{O}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{La}(\mathrm{OH})_{3}(a q)\)

Identify the unbalanced atoms

We need to balance the number of lanthanum, oxygen, and hydrogen atoms in the equation.

Balance the equation

To balance the number of lanthanum atoms, add a coefficient of 2 in front of \(\mathrm{La}(\mathrm{OH})_{3}(a q)\). Now we need to balance the number of oxygen and hydrogen atoms. There are 3 oxygen atoms in each lanthanum hydroxide unit, multiplied by 2, we have 6 oxygen atoms in total. There is also 1 oxygen atom in \(\mathrm{H}_{2} \mathrm{O}(l)\), so we do not need to change the number of oxygen atoms. But to balance the hydrogen atoms, we need to add a coefficient of 6 in front of \(\mathrm{H}_{2} \mathrm{O}(l)\). The balanced equation is: \(\mathrm{La}_{2} \mathrm{O}_{3}(s)+6\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2\mathrm{La}(\mathrm{OH})_{3}(a q)\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)

Identify the unbalanced atoms

We need to balance the number of nitrogen, oxygen, and hydrogen atoms in the equation.

Balance the equation

As there are two nitrogen atoms and four hydrogen atoms in one ammonium nitrate unit, we add a coefficient of 2 in front of \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s)\) to create a total of four nitrogen atoms in the reactants. Then, we add a coefficient of 4 in front of \(\mathrm{H}_{2} \mathrm{O}(g)\) to equalize the number of hydrogen atoms. Finally, we add a coefficient of 3 in front of \(\mathrm{O}_{2}(g)\) to balance the oxygen atoms in the products. The balanced equation is: \(2\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2}(g)+3\mathrm{O}_{2}(g)+4\mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{Ca}_{3} \mathrm{P}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{PH}_{3}(g)\)

Identify the unbalanced atoms

We need to balance the number of calcium, phosphorus, oxygen, and hydrogen atoms in the equation.

Balance the equation

To balance the calcium atoms, add a coefficient of 3 in front of \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)\). Then, to balance the phosphorus atoms, add a coefficient of 2 in front of \(\mathrm{PH}_{3}(g)\). Finally, add a coefficient of 6 in front of \(\mathrm{H}_{2} \mathrm{O}(l)\) to balance the oxygen and hydrogen atoms in the products. The balanced equation is: \(\mathrm{Ca}_{3} \mathrm{P}_{2}(s)+6\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3\mathrm{Ca}(\mathrm{OH})_{2}(a q)+2\mathrm{PH}_{3}(g)\) (e) \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{H}_{2}\mathrm{O}(l)\)

Identify the unbalanced atoms

We need to balance the number of calcium, phosphorus, oxygen, and hydrogen atoms in the equation.

Balance the equation

To balance the calcium atoms, add a coefficient of 3 in front of \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)\). Then, to balance the phosphorus atoms, add a coefficient of 2 in front of \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)\). Finally, add a coefficient of 6 in front of \(\mathrm{H}_{2}\mathrm{O}(l)\) to balance the oxygen and hydrogen atoms in the products. The balanced equation is: \(3\mathrm{Ca}(\mathrm{OH})_{2}(a q)+2\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+6\mathrm{H}_{2}\mathrm{O}(l)\) (f) \(\mathrm{AgNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Ag}_{2} \mathrm{SO}_{4}(s)+\mathrm{NaNO}_{3}(a q)\)

Identify the unbalanced atoms

We need to balance the number of silver, sodium, nitrogen, oxygen, and sulfur atoms in the equation.

Balance the equation

Add a coefficient of 2 in front of \(\mathrm{AgNO}_{3}(a q)\) to balance the silver atoms, and add a coefficient of 2 in front of \(\mathrm{NaNO}_{3}(a q)\) to balance the sodium and nitrogen atoms. The balanced equation is: \(2\mathrm{AgNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Ag}_{2} \mathrm{SO}_{4}(s)+2\mathrm{NaNO}_{3}(a q)\) (g) \(\mathrm{CH}_{3} \mathrm{NH}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)\)

Identify the unbalanced atoms

We need to balance the number of carbon, hydrogen, oxygen, and nitrogen atoms in the equation.

Balance the equation

To balance the hydrogen atoms, add a coefficient of 2 in front of \(\mathrm{H}_{2} \mathrm{O}(g)\). Then, add a coefficient of 3 in front of \(\mathrm{O}_{2}(g)\) to balance the oxygen atoms. The balanced equation is: \(\mathrm{CH}_{3} \mathrm{NH}_{2}(\mathrm{~g})+3\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)\)

Key concepts

Chemical Reactions

Chemical reactions involve substances changing into new substances. The atoms involved are rearranged, but they are neither created nor destroyed. This is the foundation of all chemical processes. Each chemical reaction can be represented by a chemical equation, which provides a clear and concise way to describe the transformation of reactants into products.

• **Reactants and Products:** Reactants are substances that start the reaction, and products are substances that result from the reaction. Their symbolic representation reflects how molecules interact and transform.
• **Chemical Equation:** This is a formulaic way to represent chemical reactions, using symbols and numbers to show how reactants are converted into products.

Understanding chemical reactions involves recognizing the specific ways in which reactant molecules collide and rearrange to form new products. This process is critical in fields ranging from basic biochemistry to complex industrial chemistry.

Stoichiometry

Stoichiometry is the study of the quantitative relationships in chemical reactions. It allows us to calculate how much of each reactant we need to produce a desired amount of product, or how much product can be formed from a given amount of reactants.

• **Moles and Mole Ratios:** The concept of moles is central to stoichiometry. Balanced chemical equations provide the mole ratio of reactants and products, allowing chemists to precisely calculate and predict reaction outcomes.
• **Conversion Factors:** Stoichiometry involves converting between units of mass, moles, and molecules, based on the relationships established in the chemical equation. This ensures proper scaling in both laboratory and industrial processes.

Stoichiometry is essential for understanding the efficiency of chemical reactions, minimizing waste, and properly scaling reactions for practical applications.

Chemical Equation Coefficients

In chemical equations, coefficients indicate the number of moles of each substance involved in the reaction. These coefficients are crucial for balancing chemical equations, ensuring the same number of each type of atom on both sides of the equation.

• **Balancing Equations:** This process involves adjusting the coefficients of reactants and products to conserve mass, following the law of conservation of mass.
• **Significance of Coefficients:** Coefficients provide the mole ratio necessary for calculations in stoichiometry. They help in predicting how much of each substance is needed or produced.

Properly balanced equations are foundational to accurate stoichiometric calculations and successful chemical reactions.

Conservation of Mass

The law of conservation of mass states that mass cannot be created or destroyed in a chemical reaction. This principle is vital when balancing chemical equations, as it assures that mass remains consistent before and after the reaction.

• **Consistent Mass:** Before a reaction and after, the mass is the same, which means that the number of atoms of each element is conserved.
• **Importance for Calculations:** Preparing balanced equations requires taking into account the conservation of mass, ensuring accurate predictions and calculations.

This principle ensures the predictability and reproducibility of chemical reactions, underpinning all experimental and industrial chemical processes.