Calculate the moles of \(\mathrm{H}_{2} \mathrm{O}\) vapours formed if \(1.57\) mole of \(\mathrm{O}_{2}\) are used in presence of excess of \(\mathrm{H}_{2}\) for the given change, $$ 2 \mathrm{H}_{2}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} $$

Understanding how to balance chemical equations is critical for solving stoichiometry problems. When we look at the reaction between hydrogen (\textbf{H}\(_2\)) and oxygen (\textbf{O}\(_2\)) to form water (\textbf{H}\(_2\)\textbf{O}), it's essential to ensure that the number of atoms of each element is conserved throughout the reaction. This law, known as the Law of Conservation of Mass, dictates that atoms cannot be created or destroyed in a chemical reaction.

In our exercise, the balanced chemical equation is presented as \(2\textbf{H}_2 + \textbf{O}_2 \rightarrow 2\textbf{H}_2\textbf{O}\). It shows that two molecules of hydrogen gas combine with one molecule of oxygen gas to produce two molecules of water vapor. To balance this equation, we look at the number of hydrogen and oxygen atoms on both sides. Initially, we have four hydrogen atoms and two oxygen atoms as reactants and end up with the same count in the product. This shows that the equation is properly balanced and is ready to be used for subsequent stoichiometry calculations.

The mole concept is a bridge between the microscopic world of atoms and molecules and the macroscopic world we observe. It allows us to count particles by weighing them. One mole is defined as the quantity of a substance that contains as many particles (atoms, ions, molecules) as there are atoms in 12 grams of carbon-12. This amount is called Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles per mole.

In stoichiometry problems, like the one given in our exercise, the mole provides a consistent method to convert between mass and number of particles. When the problem states that we have 1.57 moles of oxygen, it's saying we essentially have \(1.57 \times 6.022 \times 10^{23}\) molecules of oxygen. Knowing this helps us to envision how much of a substance we have and to perform calculations that involve chemical reactions, as we directly use this quantity in the mole ratio calculations.

Mole ratio calculations are fundamental to solving stoichiometry problems. These calculations come directly from the coefficients of a balanced chemical equation, representing the proportional relationship between reactants and products.

The mole ratio is used to convert moles of one substance into moles of another substance involved in the same reaction. If we refer back to our exercise, the balanced equation provided a mole ratio of 1:2 between oxygen (\textbf{O}\(_2\)) and water (\textbf{H}\(_2\)\textbf{O}). Such a ratio conveys that for every one mole of oxygen consumed, two moles of water are produced.

Here, by knowing that we have 1.57 moles of oxygen, we can utilize the mole ratio to find out the moles of water produced. Multiplying 1.57 moles of oxygen by the ratio \(\frac{2 \text{ moles of } \textbf{H}_2\textbf{O}}{1 \text{ mole of } \textbf{O}_2}\) gives us exactly 3.14 moles of water. This streamlined method of using mole ratios is invaluable for students studying chemistry and facing similar stoichiometry challenges.