In the reaction shown below, 64 \(\mathrm{g} \mathrm{CaC}_{2}\) is reacted with 64 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) $$\mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)+\mathrm{Ca}(\mathrm{OH})_{2}(s)$$ \begin{equation}\begin{array}{l}{\text { a. Which is the excess reactant, and which is }} \\ {\text { the limiting reactant? }} \\ {\text { b. What is the theoretical yield of } C_{2} \mathrm{H}_{2} ?} \\ {\text { c. What is the theoretical yield of } \mathrm{Ca}(\mathrm{OH})_{2} ?}\end{array}\end{equation}

In every chemical reaction, the limiting reactant is the substance that is completely consumed first and thus determines the maximum amount of product that can be formed. Understanding the concept of the limiting reactant is essential for predicting the quantities of products formed in a reaction.

When we convert the mass of each reactant to moles, we can compare these values to the stoichiometry of the balanced chemical equation to identify the limiting reactant. For instance, in the provided exercise, 64 grams of calcium carbide (CaC₂) reacts with 64 grams of water (H₂O). When converted to moles, we find that we have 1 mole of CaC₂ and 3.55 moles of H₂O. According to the chemical equation, 1 mole of CaC₂ requires 2 moles of H₂O to react completely. As we have more than enough H₂O, CaC₂ is our limiting reactant. Identifying the limiting reactant is crucial as it allows for the prediction of the maximum theoretical yield of the products.

The theoretical yield of a chemical reaction is the maximum amount of product that can be obtained if all the limiting reactant is used up with no losses. It is a vital concept in stoichiometry because it sets the benchmark for the efficiency of the reaction. Calculations of theoretical yield are based on the stoichiometry of the balanced chemical equation and the amount of the limiting reactant.

For the given problem, the theoretical yield is determined by the amount of CaC₂, which is the limiting reactant. Since the balanced equation shows a 1:1 ratio between CaC₂ and both products C₂H₂ and Ca(OH)₂, one mole of CaC₂ will produce one mole of each product. Remember, even though the reaction may be perfectly balanced on paper, in real life conditions, the actual yield is usually less due to various factors like incomplete reactions, side reactions, or loss of product during recovery.

The molar mass is the mass of one mole of a substance. In more practical terms, it represents the mass of a given substance (elements or compounds) that contains the same number of particles as there are atoms in 12 grams of carbon-12. The units for molar mass are g/mol, and it is a crucial bridge connecting mass (in grams) to moles, which allows chemists to convert from the macroscopic scale of grams to the molecular scale of moles.

Molar masses are used extensively in chemistry to calculate the number of moles of a substance when given the mass, as seen in the exercise. For instance, the molar mass of calcium carbide (CaC₂) is used to convert the given 64 grams to moles, which is a necessary step in solving stoichiometry problems. Knowing the molar mass of a substance aids significantly in various calculations in chemical reactions, from determining the limiting reactant to calculating the theoretical yield.