Question

Complete and balance the following equations: (a) \(\mathrm{Mg}_{3} \mathrm{~N}_{2}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (b) \(\mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow\) (c) \(\mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (d) \(\mathrm{NH}_{3}(a q)+\mathrm{H}^{+}(a q) \longrightarrow\) (e) \(\mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow\)

Short answer

The short answers for the balanced equations are: (a) \( 3 \ \mathrm{Mg}_{3} \mathrm{N}_{2}(\mathrm{s})+6 \ \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 6 \ \mathrm{Mg(OH)}_{2} \mathrm{~(s)} + 6 \ \mathrm{NH}_{3} \mathrm{(g)} \) (b) \( 2 \ \mathrm{NO}(g) + \mathrm{O}_{2}(g) \longrightarrow 2 \ \mathrm{NO}_{2}(g) \) (c) \( \mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \ \mathrm{HNO}_{3} \mathrm{(a~q)} \) (d) \( \mathrm{NH}_{3}(a~q)+\mathrm{H}^{+}(a~q) \longrightarrow \mathrm{NH}_{4}^{+}(a~q) \) (e) \( \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g) + 2 \ \mathrm{H}_{2} \mathrm{O}(l) \)

Step-by-step solution

a) Balancing the equation for Mg₃N₂(s) + H₂O(l) ⟶

First, identify the reactants and products in the given equation: Reactants: Mg₃N₂(s) and H₂O(l) Products: Unknown, but since it involves magnesium and nitrogen, the products would be Mg(OH)₂ and NH₃. So the balanced equation should look like: \[ \mathrm{Mg}_{3} \mathrm{N}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Mg(OH)}_{2} \mathrm{~(s)} + \mathrm{NH}_{3} \mathrm{(g)} \] Now, count the number of atoms for each element on both sides: Reactants: 3 Mg, 2 N, and 2 H Products: 1 Mg, 1 N, 4 H To balance the equation, add coefficients in front of the chemical formulas: \( 3 \ \mathrm{Mg}_{3} \mathrm{N}_{2}(\mathrm{s})+6 \ \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 6 \ \mathrm{Mg(OH)}_{2} \mathrm{~(s)} + 6 \ \mathrm{NH}_{3} \mathrm{(g)} \) The balanced equation is: \( 3 \ \mathrm{Mg}_{3} \mathrm{N}_{2}(\mathrm{s})+6 \ \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 6 \ \mathrm{Mg(OH)}_{2} \mathrm{~(s)} + 6 \ \mathrm{NH}_{3} \mathrm{(g)} \)

b) Balancing the equation for NO(g) + O₂(g) ⟶

First, identify the reactants and products in the given equation: Reactants: NO(g) and O₂(g) Products: Unknown, but since it involves nitrogen and oxygen, the product would be NO₂. So the balanced equation should look like: \( \mathrm{NO}(g) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) \) Now, count the number of atoms for each element on both sides: Reactants: 1 N and 3 O Products: 1 N and 2 O To balance the equation, add coefficients in front of the chemical formulas: \( 2 \ \mathrm{NO}(g) + \mathrm{O}_{2}(g) \longrightarrow 2 \ \mathrm{NO}_{2}(g) \) The balanced equation is: \( 2 \ \mathrm{NO}(g) + \mathrm{O}_{2}(g) \longrightarrow 2 \ \mathrm{NO}_{2}(g) \)

c) Balancing the equation for N₂O₅(g) + H₂O(l) ⟶

First, identify the reactants and products in the given equation: Reactants: N₂O₅(g) and H₂O(l) Products: Unknown, but since it involves nitrogen, oxygen, and hydrogen, the product would be HNO₃. So the balanced equation should look like: \[ \mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HNO}_{3} \mathrm{(a~q)} \] Now, count the number of atoms for each element on both sides: Reactants: 2 N, 5 O, and 2 H Products: 1 N, 3 O, and 1 H To balance the equation, add coefficients in front of the chemical formulas: \[ \mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \ \mathrm{HNO}_{3} \mathrm{(a~q)} \] The balanced equation is: \[ \mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \ \mathrm{HNO}_{3} \mathrm{(a~q)} \]

d) Balancing the equation for NH₃(aq) + H⁺(aq) ⟶

First, identify the reactants and products in the given equation: Reactants: NH₃(aq) and H⁺(aq) Products: Unknown, but since it involves nitrogen and hydrogen, the product would be NH₄⁺. So the balanced equation should look like: \[ \mathrm{NH}_{3}(a~q)+\mathrm{H}^{+}(a~q) \longrightarrow \mathrm{NH}_{4}^{+}(a~q) \] Now, count the number of atoms for each element on both sides: Reactants: 1 N and 4 H Products: 1 N and 4 H The equation is already balanced: \[ \mathrm{NH}_{3}(a~q)+\mathrm{H}^{+}(a~q) \longrightarrow \mathrm{NH}_{4}^{+}(a~q) \]

e) Balancing the equation for N₂H₄(l) + O₂(g) ⟶

First, identify the reactants and products in the given equation: Reactants: N₂H₄(l) and O₂(g) Products: Unknown, but since it involves nitrogen, hydrogen, and oxygen, the products would be N₂ and H₂O. So the balanced equation should look like: \[ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l) \] Now, count the number of atoms for each element on both sides: Reactants: 2 N, 4 H, and 2 O Products: 2 N and 2 H To balance the equation, add coefficients in front of the chemical formulas: \[ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g) + 2 \ \mathrm{H}_{2} \mathrm{O}(l) \] The balanced equation is: \[ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g) + 2 \ \mathrm{H}_{2} \mathrm{O}(l) \]

Key concepts

Balancing Equations

Chemical reactions are balanced to ensure that the number of atoms for each element is the same on both sides of the equation. This reflects the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. Balancing equations might sound tricky, but with a few steps, you can handle them easily.

When balancing, start by identifying the reactants (the substances you start with) and the products (the substances you end up with). Write down the unbalanced equation and count the atoms of each element present on each side. Use coefficients, which are numbers placed before compounds, to adjust the number of atoms in each molecule until the equation is balanced.

For example, in the equation:

• Reactants: \( Mg_3N_2(s) + H_2O(l) \)
• Products: \( Mg(OH)_2(s) + NH_3(g) \)

You begin by writing the number of magnesium, nitrogen, and hydrogen atoms. Adjust the coefficients to have the same count of each element on both sides of the equation. Patience and practice will make this easier over time!

Stoichiometry

Stoichiometry is the part of chemistry that tells us the quantitative relationships of reactants and products in a chemical reaction. It's like a recipe for a reaction, detailing how much of each ingredient (reactant) is needed to make a certain amount of product.

The key to stoichiometry is the balanced equation. Once an equation is balanced, it can show the ratio of moles of reactants and products. This helps in calculating the amounts of substances consumed and generated in a reaction. For instance, if you know the amount of one reactant, stoichiometry can tell you how much of another reactant is needed, or how much product will form.

Consider the reaction \( 2 \, NO(g) + O_2(g) \rightarrow 2 \, NO_2(g) \):

• The balanced equation provides a ratio of 2:1:2.
• This means 2 moles of NO react with 1 mole of O₂ to produce 2 moles of NO₂.

Stoichiometry helps in understanding the proportions of each compound required and allows chemists to predict the outcomes precisely.

Inorganic Chemistry

Inorganic chemistry focuses on compounds that are not covered by organic chemistry, typically those that aren't made primarily of carbon and hydrogen. This branch of chemistry is exceptionally diverse, covering a broad category of compounds including salts, minerals, and metals.

In the context of our exercise, inorganic reactions involve compounds like magnesium nitride, nitric oxide, and hydrazine. These reactions often result in the formation of other inorganic substances. For example, we use Mg₃N₂ in a reaction to form Mg(OH)₂ and NH₃, both of which are common inorganic compounds.

Reactions in inorganic chemistry can vary greatly:

• Oxidation and reduction
• Acid-base reactions
• Precipitation reactions

The reactions demonstrate a wide range of behaviors and uses in chemistry, from industrial applications to biological significance. Inorganic chemistry provides insight into the simple yet profound transformations of substances.