Question

For each of the following unbalanced chemical equations, suppose \(25.0 \mathrm{g}\) of each reactant is taken. Show by calculation which reactant is limiting. Calculate the theoretical yield in grams of the product in boldface. a. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \rightarrow \mathbf{C} \mathbf{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) b. \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow \mathbf{N} \mathbf{O}(g)\) c. \(\mathrm{NaClO}_{2}(a q)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{ClO}_{2}(g)+\mathbf{N a C l}(a q)\) d. \(\mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightarrow \mathbf{N} \mathbf{H}_{3}(g)\)

Short answer

The theoretical yields for each reaction are as follows: a. 47.71 g of \(CO_{2}\) b. 46.88 g of \(NO\) c. 16.16 g of \(NaCl\) d. 30.34 g of \(NH_{3}\)

Step-by-step solution

Balance the equation

\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\frac{3}{2}\mathrm{O}_{2}(g) \rightarrow 2\mathbf{C} \mathbf{O}_{2}(g)+3\mathrm{H}_{2} \mathrm{O}(l)\)

Identify the limiting reactant

For 25 g of each reactant: Moles of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}=\frac{25 \text{ g}}{46.07 \text{ g/mol}}=0.542 \text{ mol}\) Moles of \(\mathrm{O}_{2}=\frac{25 \text{ g}}{32.0 \text{ g/mol}}=0.7812 \text{ mol}\) As per the balanced equation, 1 mole of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) reacts with \(\frac{3}{2}\) mole of \(\mathrm{O}_{2}\). Thus, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) is limiting, as its amount is less than the required moles of \(\mathrm{O}_{2}\).

Calculate the theoretical yield in grams

From the balanced equation, 1 mol of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) forms 2 moles of \(CO_{2}\). So moles of \(CO_{2}\) formed = 2 × 0.542 mol = 1.084 mol Theoretical yield of \(CO_{2}\): \(1.084 \text{ mol} \times 44.01 \text{ g/mol} = 47.71 \text{ g} \) b. Reaction: \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow \mathbf{N} \mathbf{O}(g)\)

Balance the equation

\(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathbf{N} \mathbf{O}(g)\)

Identify the limiting reactant

Moles of \(\mathrm{N}_{2}=\frac{25 \text{ g}}{28.02 \text{ g/mol}}=0.891 \text{ mol}\) Moles of \(\mathrm{O}_{2}=\frac{25 \text{ g}}{32.0 \text{ g/mol}}=0.7812 \text{ mol}\) As per the balanced equation, 1 mole of \(\mathrm{N}_{2}\) reacts with 1 mole of \(\mathrm{O}_{2}\). Thus, \(\mathrm{O}_{2}\) is limiting.

Calculate the theoretical yield in grams

1 mol of \(\mathrm{O}_{2}\) forms 2 moles of \(NO\). So moles of \(NO\) formed = 2 × 0.7812 mol = 1.5624 mol Theoretical yield of \(NO\): \(1.5624 \text{ mol} \times 30.01 \text{ g/mol} = 46.88 \text{ g} \) c. Reaction: \(\mathrm{NaClO}_{2}(a q)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{ClO}_{2}(g)+\mathbf{N a C l}(a q)\)

Balance the equation

\(2\mathrm{NaClO}_{2}(a q)+\mathrm{Cl}_{2}(g) \rightarrow 2\mathrm{ClO}_{2}(g)+2\mathbf{N a C l}(a q)\)

Identify the limiting reactant

Moles of \(\mathrm{NaClO}_{2}=\frac{25 \text{ g}}{90.44 \text{ g/mol}}=0.2765 \text{ mol}\) Moles of \(\mathrm{Cl}_{2}=\frac{25 \text{ g}}{70.9 \text{ g/mol}}=0.3525 \text{ mol}\) As per the balanced equation, 2 moles of \(\mathrm{NaClO}_{2}\) reacts with 1 mole of \(\mathrm{Cl}_{2}\). Thus, \(\mathrm{NaClO}_{2}\) is limiting.

Calculate the theoretical yield in grams

2 moles of \(\mathrm{NaClO}_{2}\) forms 2 moles of \(NaCl\). So moles of \(NaCl\) formed = 0.2765 mol Theoretical yield of \(NaCl\): \(0.2765 \text{ mol} \times 58.44 \text{ g/mol} = 16.16 \text{ g} \) d. Reaction: \(\mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightarrow \mathbf{N} \mathbf{H}_{3}(g)\)

Balance the equation

\(3\mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightarrow 2 \mathbf{N} \mathbf{H}_{3}(g)\)

Identify the limiting reactant

Moles of \(\mathrm{H}_{2}=\frac{25 \text{ g}}{2.02 \text{ g/mol}}=12.38 \text{ mol}\) Moles of \(\mathrm{N}_{2}=\frac{25 \text{ g}}{28.02 \text{ g/mol}}=0.891 \text{ mol}\) As per the balanced equation, 3 moles of \(\mathrm{H}_{2}\) react with 1 mole of \(\mathrm{N}_{2}\). Thus, \(\mathrm{N}_{2}\) is limiting.

Calculate the theoretical yield in grams

1 mole of \(\mathrm{N}_{2}\) forms 2 moles of \(NH_{3}\). So moles of \(NH_{3}\) formed = 2 × 0.891 mol = 1.782 mol Theoretical yield of \(NH_{3}\): \(1.782 \text{ mol} \times 17.03 \text{ g/mol} = 30.34 \text{ g} \)

Key concepts

Limiting Reactant

Imagine you're making sandwiches with one slice of cheese per two slices of bread. If you have 1 slice of cheese but 10 slices of bread, you can only make one sandwich. In chemistry, this is similar to the concept of the limiting reactant; it's the ingredient that runs out first and limits the amount of product you can make in a chemical reaction.

Just like the slice of cheese, the limiting reactant is the chemical that determines the maximum amount of product that can be formed. When conducting experiments or calculating theoretical yields, identifying the limiting reactant is crucial. You calculate the amount of product that can be formed by each reactant and the one that forms the least amount of product is the limiting reactant. This ensures accurate predictions and efficient resource use in real-world applications such as pharmaceuticals manufacturing or food chemistry.

Chemical Equation Balancing

Before a chemist can perform any sort of calculation, such as finding the limiting reactant, they must first make sure the chemical equation is balanced. This means adjusting the coefficients (the numbers in front of the chemical formulas) to make sure the number of atoms for each element is the same on both sides of the equation.

It's just like a seesaw; for it to balance, you need the same weight on both sides. In a chemical equation, instead of weight, we're balancing atoms. This not only obeys the law of conservation of matter but also allows for accurate stoichiometric calculations. Stoichiometry is just a fancy word for the math that chemists use to relate the quantities of reactants and products in a chemical reaction.

Mole Concept

In chemistry, we often deal with quantities that are way too large to count individually. That's where the mole concept comes in to save the day. A mole is a standard unit in chemistry that represents a massive quantity of molecules or atoms, roughly 6.022 x 10²³ particles, known as Avogadro's number.

The mole concept is akin to buying eggs by the dozen; it's just a handy batch size. Scientists use moles because it helps them relate masses of different substances in a chemical reaction. Since we can't count out billions of atoms when making reactions, we weigh them instead, and thanks to the mole concept, we know how many particles we have from that weight. This is essential in calculating theoretical yields and determining limiting reactants.