Question

Consider the reaction: $$ 4 \mathrm{~K}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{~K}_{2} \mathrm{O}(s) $$ The molar mass of \(\mathrm{K}\) is \(39.10 \mathrm{~g} / \mathrm{mol}\), and that of \(\mathrm{O}_{2}\) is \(32.00 \mathrm{~g} / \mathrm{mol}\). Without doing any calculations, pick the conditions under which potassium is the limiting reactant and explain your reasoning. a. \(170 \mathrm{~g} \mathrm{~K}, 31 \mathrm{~g} \mathrm{O}_{2}\) b. \(16 \mathrm{~g} \mathrm{~K}, 2.5 \mathrm{~g} \mathrm{O}_{2}\) c. \(165 \mathrm{~kg} \mathrm{~K}, 28 \mathrm{~kg} \mathrm{O}_{2}\) d. \(1.5 \mathrm{~g} \mathrm{~K}, 0.38 \mathrm{~g} \mathrm{O}_{2}\)

Short answer

Under condition d, potassium is the limiting reactant because the mole ratio of K to O2 is less than the required 4:1 ratio from the balanced equation.

Step-by-step solution

- Understanding the limiting reactant

The limiting reactant is the one that will be completely consumed first in a chemical reaction, determining the maximum amount of product that can be formed. To find the limiting reactant, compare the mole ratio of the reactants provided to the mole ratio required by the balanced chemical equation.

- Analyzing the mole ratio in the balanced equation

The balanced chemical equation shows that 4 moles of K react with 1 mole of O2 to produce 2 moles of K2O. This means that for every 4 moles of K, we need 1 mole of O2.

- Review each option separately

Compare the amount in moles of each reactant provided in each condition to the ratio from the balanced equation to determine if K will be the limiting reactant: a. For 170 g K, we have \( 170 \text{ g} \div (39.10 \text{ g/mol}) = 4.35 \text{ moles of K} \) For 31 g O2, we have \( 31 \text{ g} \div (32.00 \text{ g/mol}) = 0.97 \text{ moles of O2} \) In mole ratio of 4 moles of K to 1 mole of O2, K is in excess. b. For 16 g K, we have \( 16 \text{ g} \div (39.10 \text{ g/mol}) = 0.41 \text{ moles of K} \) For 2.5 g O2, we have \( 2.5 \text{ g} \div (32.00 \text{ g/mol}) = 0.08 \text{ moles of O2} \) Here, K is in excess again. c. For 165 kg K (or 165,000 g), we have \( 165,000 \text{ g} \div (39.10 \text{ g/mol}) = 4,219.44 \text{ moles of K} \) For 28 kg O2 (or 28,000 g), we have \( 28,000 \text{ g} \div (32.00 \text{ g/mol}) = 875 \text{ moles of O2} \) Once again, O2 is the limiting reactant. d. For 1.5 g K, we have \( 1.5 \text{ g} \div (39.10 \text{ g/mol}) = 0.038 \text{ moles of K} \) For 0.38 g O2, we have \( 0.38 \text{ g} \div (32.00 \text{ g/mol}) = 0.012 \text{ moles of O2} \) In this case, the mole ratio required for 4 moles of K to 1 mole of O2 is not satisfied since there is not enough K, making K the limiting reactant.

Key concepts

Chemical Reaction Stoichiometry

Understanding chemical reaction stoichiometry is essential for mastering chemistry, especially when balancing equations and predicting product formations. Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It involves calculations that allow chemists to predict the amounts of substances consumed and produced in a given reaction.

For example, suppose you're given the reaction of potassium (\text{K}) with oxygen (\text{O}_{2}) to form potassium oxide (\text{K}_{2}\text{O}). According to the balanced chemical equation, \(4 \text{K}(s) + \text{O}_2(g) \rightarrow 2 \text{K}_2\text{O}(s)\), 4 atoms of potassium react with one molecule of oxygen to produce 2 units of potassium oxide. Stoichiometry allows for the calculation of how much product can be formed from a certain amount of starting material.

It becomes particularly useful when trying to determine the 'limiting reactant' – that is, the reactant that is entirely consumed first, thereby limiting the amount of product formed. Knowing the concept behind stoichiometry can significantly aid in tasks like optimizing resources in industrial processes or conducting precise scientific experiments.

Mole Ratio

The mole ratio plays a pivot role in the study of stoichiometry as it relates the quantities of reactants and products in a balanced chemical equation. The mole is the standard unit for amount of substance in chemistry and is used to express quantities in manageable numbers. For a given balanced equation, the coefficients indicate the mole ratios required for the reaction to proceed.

In the reaction of potassium and oxygen, the balanced equation indicates a 4:1 ratio of potassium to oxygen. It signifies that 4 moles of potassium are needed to fully react with 1 mole of oxygen. It's essential to comprehend this concept because when given the masses of the reactants, we can convert them to moles and use the mole ratio to identify the limiting reactant.

For instance, by converting grams to moles, you will be able to see how many moles of each reactant you have. An imbalance in the mole ratio indicates excess of one reactant over the other. As a result, one reactant will be left over when the other is completely consumed – identifying the limiting reactant. Understanding mole ratios can assist you in predicting the amounts of products formed and designing reactions where desired products are maximized.

Theoretical Yield

The theoretical yield is the amount of product that can be produced in a chemical reaction based on the limiting reactant, assuming that the reaction goes to completion with no losses. It represents the maximum amount of product that can be generated from the given amounts of reactants.

To calculate the theoretical yield, we first determine which reactant is limiting the reaction and then use stoichiometry to predict the maximum amount of product that can be formed. This involves converting the limiting reactant’s mass to moles, applying the stoichiometric mole ratios from the balanced equation, and then converting the moles of product into grams (or any other unit of mass).

In instructional scenarios, for example, knowing how to calculate the theoretical yield allows students to see the efficiency of a reaction. A reaction's actual yield, the amount actually produced in a laboratory setting, is often less due to practical considerations such as incomplete reactions, side reactions, or loss of product during processing. Comparing the actual yield to the theoretical yield gives the percentage yield, a measure of the reaction's efficiency. Understanding theoretical yield is a fundamental aspect of chemical synthesis, process optimization, and economic considerations in the industry.