Question

Balance each of the following neutralization reactions: a. \(\mathrm{HCl}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{MgCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) b. \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{LiOH}(a q) \longrightarrow \mathrm{Li}_{3} \mathrm{PO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Short answer

a. \(2 \text{HCl} + \text{Mg}(\text{OH})_{2} \rightarrow \text{MgCl}_{2} + 2 \text{H}_{2} \text{O}\) b. \( \text{H}_{3} \text{PO}_{4} + 3 \text{LiOH} \rightarrow \text{Li}_{3} \text{PO}_{4} + 3 \text{H}_{2} \text{O} \)

Step-by-step solution

- Identify the reactants and products

For part a, the reactants are \(\text{HCl}(aq)\) and \( \text{Mg}(\text{OH})_{2}(s)\) and the products are \( \text{MgCl}_{2}(aq)\) and \( \text{H}_{2} \text{O}(l)\). For part b, the reactants are \( \text{H}_{3} \text{PO}_{4}(aq)\) and \( \text{LiOH}(aq)\) and the products are \( \text{Li}_{3} \text{PO}_{4}(aq)\) and \( \text{H}_{2} \text{O}(l)\).

- Balance the elements in part a

For part a, balance the elements as follows:\( \text{HCl}(aq) + \text{Mg}(\text{OH})_{2}(s) \rightarrow \text{MgCl}_{2}(aq) + \text{H}_{2} \text{O}(l) \).Start with magnesium (Mg), which is already balanced with 1 atom on each side.Next, balance chlorine (Cl). There are 2 chlorine atoms in \( \text{MgCl}_{2}\), so you will need 2 molecules of HCl on the reactant side: \( 2 \text{HCl}(aq) + \text{Mg}(\text{OH})_{2}(s) \rightarrow \text{MgCl}_{2}(aq) + \text{H}_{2} \text{O}(l) \).Finally, balance the hydrogen and oxygen atoms. There are 2 hydrogen atoms from HCl and 2 from \( \text{Mg}(\text{OH})_{2}\) for a total of 4 hydrogen atoms, and 2 oxygen atoms from \( \text{Mg}(\text{OH})_{2}\). Four hydrogen atoms paired with 2 oxygen atoms produce 2 water molecules: \( 2 \text{HCl}(aq) + \text{Mg}(\text{OH})_{2}(s) \rightarrow \text{MgCl}_{2}(aq) + 2 \text{H}_{2} \text{O}(l) \).

- Balance the elements in part b

For part b, balance the elements as follows:\( \text{H}_{3} \text{PO}_{4}(aq) + \text{LiOH}(aq) \rightarrow \text{Li}_{3} \text{PO}_{4}(aq) + \text{H}_{2} \text{O}(l) \).First, balance lithium (Li). There are 3 lithium atoms in \( \text{Li}_{3} \text{PO}_{4}\), so you will need 3 molecules of \( \text{LiOH} \) on the reactant side: \( \text{H}_{3} \text{PO}_{4}(aq) + 3 \text{LiOH}(aq) \rightarrow \text{Li}_{3} \text{PO}_{4}(aq) + \text{H}_{2} \text{O}(l) \).Next, balance hydrogen and oxygen atoms. There are 3 oxygen atoms from \( \text{LiOH}\) and 1 from \( \text{H}_{3} \text{PO}_{4}\) for a total of 4 oxygen atoms, which produce 3 molecules of water. Also, there are 3 hydrogen atoms from \( \text{H}_{3} \text{PO}_{4}\) and 3 from \( \text{LiOH}\) for a total of 6 hydrogen atoms, paired in 3 water molecules: \( \text{H}_{3} \text{PO}_{4}(aq) + 3 \text{LiOH}(aq) \rightarrow \text{Li}_{3} \text{PO}_{4}(aq) + 3 \text{H}_{2} \text{O}(l) \).

Key concepts

Neutralization Reactions

Neutralization reactions are a type of chemical reaction where an acid and a base react to form water and a salt. In these reactions, the hydrogen ions (H+) from the acid combine with the hydroxide ions (OH-) from the base to create water (H2O). The remaining ions form the salt. Neutralization reactions are crucial in chemistry because they help to regulate pH levels in various environments and processes.

Chemical Stoichiometry

Chemical stoichiometry focuses on the quantitative relationships between the reactants and products in a chemical reaction. The principles of stoichiometry ensure that the law of conservation of mass is obeyed, meaning the total mass of the reactants must equal the total mass of the products. Understanding stoichiometry is key to predicting the amounts of substances required and produced in a chemical reaction.
To apply stoichiometry, we use balanced chemical equations. A balanced equation has equal numbers of each type of atom on both sides of the reaction arrow. This balance reflects the conservation of matter and ensures that all atoms present in the reactants are accounted for in the products.

Balancing Equations Step-by-Step

Balancing chemical equations is a fundamental skill in chemistry. Here is a simple guide to help with the process:
1. **Write the unbalanced equation:** List the reactants and the products.
2. **Count the atoms:** Make a list of how many of each type of atom are on both sides of the equation.
3. **Balance one element at a time:** Start with elements that appear only once on each side. Adjust coefficients to balance the number of atoms.
4. **Check the balance:** Ensure that the number of atoms for each element is equal on both sides.
5. **Repeat if necessary:** If any elements are still unbalanced, adjust their coefficients as needed.
For example, in the given exercise, we balance the reaction step-by-step. For reaction (a) of HCl and Mg(OH)2, we add coefficients to achieve:2 HCl + Mg(OH)2 → MgCl2 + 2 H2O
For reaction (b) of H3PO4 and LiOH, we balance it to get:H3PO4 + 3 LiOH → Li3PO4 + 3 H2O
This ensures that each element has the same number of atoms on both sides of the equation.