Question

Balance each of the following equations: a. \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)\) b. \(\mathrm{HgO}(s) \longrightarrow \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) c. \(\mathrm{Fe}(s)+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) d. \(\mathrm{Na}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NaCl}(s)\)

Short answer

a. \ \(\mathrm{N_{2} + O_{2} \longrightarrow 2NO}\)\ b. \ \(\mathrm{2HgO \longrightarrow 2Hg + O_{2}}\)\ c. \ \(\mathrm{4Fe + 3O_{2} \longrightarrow 2Fe_{2}O_{3}}\)\ d. \ \(\mathrm{2Na + Cl_{2} \longrightarrow 2NaCl}\)

Step-by-step solution

Write down the unbalanced equation

a. \ \(\mathrm{N_{2}(g) + O_{2}(g) \longrightarrow NO(g)}\)\ b. \ \(\mathrm{HgO(s) \longrightarrow Hg(l) + O_{2}(g)}\)\ c. \ \(\mathrm{Fe(s) + O_{2}(g) \longrightarrow Fe_{2}O_{3}(s)}\)\ d. \ \(\mathrm{Na(s) + Cl_{2}(g) \longrightarrow NaCl(s)}\)

Balance equation (a)

Count atoms for each element:\ On the left: 2 Nitrogen (N), 2 Oxygen (O)\ On the right: 1 Nitrogen (N), 1 Oxygen (O)\ To balance, place coefficient 2 in front of NO:\ \(\mathrm{N_{2}(g) + O_{2}(g) \longrightarrow 2NO(g)}\)

Balance equation (b)

Count atoms for each element:\ On the left: 1 Mercury (Hg), 1 Oxygen (O)\ On the right: 1 Mercury (Hg), 2 Oxygen (O)\ To balance, place coefficient 2 in front of HgO:\ \(\mathrm{2HgO(s) \longrightarrow 2Hg(l) + O_{2}(g)}\)

Balance equation (c)

Count atoms for each element:\ On the left: 1 Iron (Fe), 2 Oxygen (O)\ On the right: 2 Iron (Fe), 3 Oxygen (O)\ To balance, place coefficient 4 in front of Fe and 3 in front of O2:\ \(\mathrm{4Fe(s) + 3O_{2}(g) \longrightarrow 2Fe_{2}O_{3}(s)}\)

Balance equation (d)

Count atoms for each element:\ On the left: 1 Sodium (Na), 2 Chlorine (Cl)\ On the right: 1 Sodium (Na), 1 Chlorine (Cl)\ To balance, place coefficient 2 in front of Na and NaCl:\ \(\mathrm{2Na(s) + Cl_{2}(g) \longrightarrow 2NaCl(s)}\)

Key concepts

stoichiometry

Stoichiometry is a key concept in chemistry that involves calculating the quantities of reactants and products in chemical reactions. It helps us understand how reactants transform into products and ensures that we use the right proportions. In the context of balancing chemical equations:

• We determine the mole ratio of reactants and products.
• This ratio helps us balance the equation by ensuring that the number of atoms for each element is the same on both sides of the reaction.
• This concept is vital for predicting yields and scaling reactions appropriately.

law of conservation of mass

The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. This principle is fundamental for balancing chemical equations as it ensures that:

• The total mass of reactants equals the total mass of products.
• Each element must have the same number of atoms on both sides of the equation.
• This law helps us verify that a chemical equation is balanced and accurate.

When applying this law, we make sure to account for every atom involved in the reaction to maintain the balance.

chemical reactions

A chemical reaction involves the transformation of reactants into products through the breaking and forming of chemical bonds. Understanding chemical reactions includes:

• Identifying the reactants and products.
• Writing the chemical equation that represents the reaction.
• Recognizing different types of reactions, such as synthesis, decomposition, and combustion.

For example, in the given exercise:

• In the reaction \(\text{N}_{2}(g) + \text{O}_{2}(g) \rightarrow \text{NO}(g)\), nitrogen and oxygen react to form nitric oxide.
• In \(\text{HgO}(s) \rightarrow \text{Hg}(l) + \text{O}_{2}(g)\), mercuric oxide decomposes to mercury and oxygen gas.

This understanding helps to visualize and predict the outcome of chemical processes.

coefficient adjustment

Adjusting coefficients in a chemical equation is crucial for balancing the equation correctly. Coefficients are the numbers placed before the chemical formulas to indicate the quantity of each substance involved. To successfully adjust coefficients:

• First, write down the unbalanced equation.
• Count the number of atoms for each element on both sides of the equation.
• Adjust the coefficients to ensure equal numbers of atoms for each element.

For instance, in the reaction \(\text{4Fe}(s) + \text{3O}_{2}(g) \rightarrow \text{2Fe}_{2}\text{O}_{3}(s)\):

• We place the coefficient 4 before iron (Fe) and 3 before oxygen (\text{O}_{2}).
• These adjustments balance the number of iron and oxygen atoms on both sides.

This systematic approach ensures that the equation respects the law of conservation of mass and accurately represents the chemical reaction.