When \(56.6 \mathrm{~g}\) of calcium and \(30.5 \mathrm{~g}\) of nitrogen gas undergo a reaction that has a \(93.0 \%\) yield, what mass (g) of calcium nitride forms?

To begin with any stoichiometry problem, writing the balanced chemical equation is crucial. A balanced chemical equation ensures that the law of conservation of mass is upheld by having the same number of atoms of each element on both sides of the equation. For the reaction between calcium (\text{Ca}) and nitrogen gas (\text{N}_2), the balanced equation is:

\[ 3\text{Ca} + \text{N}_2 \rightarrow \text{Ca}_3\text{N}_2 \]

This tells us that 3 moles of calcium react with 1 mole of nitrogen gas to produce 1 mole of calcium nitride (\text{Ca}_3\text{N}_2). Without a balanced equation, calculating moles and related quantities would be inaccurate, leading to incorrect results.

The limiting reactant is the substance that is completely consumed first in a chemical reaction, thus determining the maximum amount of product that can be formed. To identify the limiting reactant, we compare the mole ratio of the reactants given in the balanced equation to the amount of reactants available.

In the problem, we have 1.412 moles of calcium and 1.089 moles of nitrogen gas. According to the equation, we need 3 moles of calcium for every 1 mole of nitrogen. Hence, the required moles of calcium for 1.089 moles of nitrogen would be:

\[ 3 \times 1.089 = 3.267 \text{ mol of Ca} \]

Since we only have 1.412 moles of calcium, which is less than 3.267, calcium is the limiting reactant. This means calcium will be completely used up first, and it determines the maximum amount of calcium nitride produced.

The theoretical yield is the maximum amount of product that can be produced from a given amount of limiting reactant, assuming the reaction goes to completion with no losses. To calculate the theoretical yield, we use the moles of the limiting reactant and the balanced equation.

In this case, the moles of calcium nitride (\text{Ca}_3\text{N}_2) produced from the limiting reactant (calcium) is calculated by:

\[ \text{Moles of Ca}_3\text{N}_2 = \frac{1.412 \text{ mol Ca}}{3} = 0.471 \text{ mol Ca}_3\text{N}_2 \]

We then use the molar mass of calcium nitride (148.25 g/mol) to find the mass:

\[ \text{Mass of Ca}_3\text{N}_2 = 0.471 \text{ mol} \times 148.25 \text{ g/mol} = 69.82 \text{ g} \]

Thus, the theoretical yield of calcium nitride is 69.82 grams.

The percent yield indicates the efficiency of a reaction by comparing the actual yield (the amount of product actually obtained from the reaction) to the theoretical yield. It is calculated using the formula:

\[ \text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100 \]

In our problem, we are given the percent yield is 93.0%. Using the theoretical yield we previously calculated (69.82 grams), we find the actual yield as follows:

\[ \text{Actual Yield} = 69.82 \text{ g} \times 0.93 = 64.93 \text{ g} \]

Therefore, after accounting for the percent yield, the mass of calcium nitride actually formed is 64.93 grams. Percent yield helps in understanding how close we are to achieving the theoretical maximum, considering practical limitations and losses during the reaction.