Balance the following equations: (a) \(\mathrm{Ag}(s)+\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Ag}_{2} \mathrm{S}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) (b) \(\mathrm{P}_{4}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)\) (c) \(\operatorname{Pb}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Pb}(\mathrm{OH})_{2}(s)\) (d) \(\operatorname{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{H}_{2}(g)\) (e) \(\mathrm{Sc}_{2} \mathrm{O}_{3}(s)+\mathrm{SO}_{3}(l) \longrightarrow \mathrm{Sc}_{2}\left(\mathrm{SO}_{4}\right)_{3}(s)\) (f) \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{Ca}\left(\mathrm{H}_{2} \mathrm{PO}_{4}\right)_{2}(a q)\) (g) \(\mathrm{Al}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(s)+\mathrm{H}_{2}(g)\) (h) \(\operatorname{TiCl}_{4}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{TiO}_{2}(s)+\mathrm{HCl}(g)\)

Understanding chemical reaction notation is fundamental for mastering chemistry. When we write a chemical equation, it's an abbreviated way to describe a chemical reaction. Each substance in a reaction is represented by its chemical formula, and the equation shows the relationship between reactants and products.

For example, in equation (a) from our exercise \(\mathrm{Ag}(s)+\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Ag}_{2} \mathrm{S}(s)+\mathrm{H}_{2} \mathrm{O}(l)\), \(\mathrm{Ag}(s)\), \(\mathrm{H}_{2} \mathrm{S}(g)\), and \(\mathrm{O}_{2}(g)\) are reactants, while \(\mathrm{Ag}_{2} \mathrm{S}(s)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) are products. The \((s)\), \((g)\), and \((l)\) symbols indicate the physical states of the substances: solid, gas, and liquid, respectively.

When balancing, each chemical's coefficient (the number in front) must be determined to satisfy the law of conservation of mass. Every chemical symbol and polyatomic ion must be balanced to reflect the true stoichiometry of the reaction. This process requires practice, but once you understand the basics, you'll find it easier to write and balance chemical equations correctly.

Stoichiometry is the study of the quantitative relationships or ratios between reactants and products in chemical reactions. It's like a quantitative recipe for chemistry. When balancing equations, stoichiometry ensures that the number of atoms for each element is the same on both sides of the reaction.

Look at exercise (b) \(\mathrm{P}_{4}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)\). Once the equation is balanced, the stoichiometry indicates that one molecule of tetraphosphorus reacts with five molecules of oxygen to produce one molecule of tetraphosphorus decaoxide. Stoichiometry also helps in calculating reactant or product quantities needed or produced in a reaction. If you understand stoichiometry, you can determine how much of a reactant you need to create a certain amount of product and vice versa.

In practical terms, if you know the amount of one reactant, you can calculate the amount of others required or the amount of products formed using stoichiometric coefficients and molar masses of the substances involved.

The law of conservation of mass is a fundamental principle in chemistry stating that mass is neither created nor destroyed in a chemical reaction. This means the mass of the reactants must equal the mass of the products. Balancing chemical equations respects this law by ensuring that the same number of each type of atom appears on both sides of the equation.

In our exercises, such as (g) \(\mathrm{Al}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(s)+\mathrm{H}_{2}(g)\), we can't simply change the chemical formulas to balance the equation; we must add coefficients to indicate the number of units of each compound or molecule that participate in the reaction. When we say that an equation is balanced, we affirm that the law of conservation of mass is satisfied.

The implication of this law extends beyond paper. In laboratories and industries, accurate measurements matter because the law of conservation of mass dictates how much product can be made from given reactants. Misunderstanding or miscalculating this can lead to inefficient reactions and wasted materials.