Phosphoric acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{4}\right)\) prepared in a two step process. (1) \(\mathrm{P}_{4}+5 \mathrm{O}_{2} \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}\) (2) \(\mathrm{P}_{4} \mathrm{O}_{10}+6 \mathrm{H}_{2} \mathrm{O} \longrightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4}\) We allow \(62 \mathrm{~g}\) of phosphorus to react with excess oxygen which form \(\mathrm{P}_{4} \mathrm{O}_{10}\) in \(85 \%\) yield. In the step (2) reaction \(90 \%\) yield of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is obtained. Produced mass of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is: (a) \(37.485 \mathrm{~g}\) (b) \(149.949 \mathrm{~g}\) (c) \(125.47 \mathrm{~g}\) (d) \(564.48 \mathrm{~g}\)

Understanding chemical yield is crucial when studying reactions on paper and performing them in a laboratory. In the world of chemistry, reactions are often not perfect, which means not all reactants turn into the desired products. The percentage of the theoretical amount of product that is actually produced by a chemical reaction is referred to as the percentage yield.

To calculate the actual yield, we start with the theoretical yield, which is the maximum amount of product that could be formed from the given amounts of reactants. This is based on the stoichiometry of the balanced chemical equation. However, due to factors like incomplete reactions, side reactions, and loss of product during recovery, the actual yield is often less than the theoretical yield. We express the efficiency of a reaction in terms of percentage yield, calculated using the formula:\[\begin{equation}\text{Percentage Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\%\end{equation}\]In the given exercise, we apply this concept twice—first for the yield of \(\text{P}_4\text{O}_{10}\) and then for \(\text{H}_3\text{PO}_{4}\) in subsequent steps of the reactions.

The molar mass is a fundamental concept in chemistry, providing a bridge between the atomic scale and the day-to-day laboratory scale. It is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). One mole of any substance contains Avogadro's number of entities (particles, atoms, molecules, ions, etc.), which is approximately \(6.022 \times 10^{23}\).

To calculate molar mass, you sum the atomic masses of all the atoms in a molecule. For instance, in the exercise, \(\text{P}_4\) has a molar mass of \(4\) times the atomic mass of phosphorus (approximately 30.97 g/mol), giving us \(123.88\) g/mol. This calculation allows us to then convert between grams and moles—a critical step in solving stoichiometry problems.Molar Mass Calculation for \(\text{H}_3\text{PO}_{4}\):\[\text{Molar Mass of } \text{H}_3\text{PO}_{4} = 3(\text{H}) + \text{P} + 4(\text{O})\]\[\text{Molar Mass of } \text{H}_3\text{PO}_{4} = 3(1.01\text{ g/mol}) + 30.97\text{ g/mol} + 4(16.00\text{ g/mol})\]\[\text{Molar Mass of } \text{H}_3\text{PO}_{4} = 98.00\text{ g/mol}\]This value is used in converting the moles of \(\text{H}_3\text{PO}_{4}\) obtained from the stoichiometry to grams for the final answer.

The mole concept is a method of quantifying particles, like atoms and molecules, by using a fixed number (Avogadro's number \(6.022 \times 10^{23}\)). It allows chemists to count particles by weighing, as counting each particle individually is impossible. With the mole concept, if you have one mole of any substance, you have \(6.022 \times 10^{23}\) units of that substance, regardless of what it is.

The mole concept connects the micro world of atoms and molecules to the macro world we can measure, which is fundamental in stoichiometry problems. For example, in the exercise, we first calculate the moles of \(P_4\) using its molar mass and the mass given. Then, we use the mole ratios from the balanced chemical equations to calculate how many moles of products are expected (the theoretical yield) and adjust for the percentage yield to find out how many moles were actually produced (the actual yield).

The entire series of calculations from mass to moles, moles to moles (using the stoichiometric coefficients), and back to mass (after adjusting for yield) is rooted in the mole concept and showcases its vital role in solving stoichiometry problems.