Question

Consider the reaction $$ 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) $$ Identify the limiting reagent in each of the reaction mixtures given below: a. 50 molecules of \(\mathrm{H}_{2}\) and 25 molecules of \(\mathrm{O}_{2}\) b. 100 molecules of \(\mathrm{H}_{2}\) and 40 molecules of \(\mathrm{O}_{2}\) c. 100 molecules of \(\mathrm{H}_{2}\) and 100 molecules of \(\mathrm{O}_{2}\) d. \(0.50 \mathrm{~mol} \mathrm{H}_{2}\) and \(0.75 \mathrm{~mol} \mathrm{O}\). e. \(0.80 \mathrm{~mol} \mathrm{H}_{2}\) and \(0.75 \mathrm{~mol} \mathrm{O}_{2}\) f. \(1.0 \mathrm{~g} \mathrm{H}_{2}\) and \(0.25 \mathrm{~mol} \mathrm{O}_{2}\) g. \(5.00 \mathrm{~g} \mathrm{H}_{2}\) and \(56.00 \mathrm{~g} \mathrm{O}_{2}\)

Short answer

a. Limiting reagent: \( \mathrm{H}_{2} \) b. Limiting reagent: \( \mathrm{O}_{2} \) c. Limiting reagent: \( \mathrm{H}_{2} \) d. Limiting reagent: \( \mathrm{H}_{2} \) e. Limiting reagent: \( \mathrm{H}_{2} \) f. No limiting reagent g. Limiting reagent: \( \mathrm{H}_{2} \)

Step-by-step solution

Identify the given Molecules/Moles

We are given a range of different scenarios with different amounts of molecules and moles of the reactants. For each scenario, we will compare the given ratio of \(\mathrm{H}_2\) and \(\mathrm{O}_2\) to the stoichiometric ratio in the balanced equation.

Compare Ratios

For each scenario, we'll calculate the ratios of the given reactants to the needed stoichiometric ratio and identify the limiting reagent. a. For 50 molecules of \( \mathrm{H}_{2} \) and 25 molecules of \( \mathrm{O}_{2} \): We can see that there are enough molecules of \( \mathrm{O}_{2} \) present for the reaction to proceed entirely. Therefore, the limiting reagent in this case is \( \mathrm{H}_{2} \). b. For 100 molecules of \( \mathrm{H}_{2} \) and 40 molecules of \( \mathrm{O}_{2} \): Divide the amount of molecules by the stoichiometric coefficients (2 for \( \mathrm{H}_{2} \) and 1 for \( \mathrm{O}_{2} \)): \[\frac{100}{2} : \frac{40}{1} = 50:40\] Since there are more available moles of \( \mathrm{H}_{2} \) compared to the stoichiometric ratio, the limiting reagent is \( \mathrm{O}_{2} \). c. For 100 molecules of \( \mathrm{H}_{2} \) and 100 molecules of \( \mathrm{O}_{2} \): Divide the molecules by the stoichiometric coefficients to get a ratio of 50:100. Thus, the limiting reagent in this case is \( \mathrm{H}_{2} \). d. For \( 0.50 \ \mathrm{mol} \ \mathrm{H}_{2} \) and \( 0.75 \ \mathrm{mol} \ \mathrm{O}_{2} \): Calculate the ratio of the moles: \[\frac{0.50}{2}: \frac{0.75}{1} = 0.25:0.75\] Since there are fewer available moles of \(\mathrm{H}_2\) as compared to the stoichiometric ratio, the limiting reagent is \( \mathrm{H}_{2} \). e. For \( 0.80 \ \mathrm{mol} \ \mathrm{H}_{2} \) and \( 0.75 \ \mathrm{mol} \ \mathrm{O}_{2} \): Calculate the ratio of the moles: \[\frac{0.80}{2}: \frac{0.75}{1} = 0.40:0.75\] Since there are fewer available moles of \(\mathrm{H}_2\) as compared to the stoichiometric ratio, the limiting reagent is \( \mathrm{H}_{2} \). f. For \((1.0 \ \mathrm{g} \ \mathrm{H}_{2})\) and \((0.25 \ \mathrm{mol} \ \mathrm{O}_{2})\): First, convert the grams of \( \mathrm{H}_{2} \) to moles (molar mass of \( \mathrm{H}_{2} = 2 \ \mathrm{g/mol} \)): \[\frac{1.0 \mathrm{~g} \mathrm{H}_{2}}{2 \mathrm{~g} \mathrm{H}_{2 / mol}} = 0.50 \ \mathrm{mol} \ \mathrm{H}_{2} \] Now calculate the ratio of the moles: \[\frac{0.50}{2}: \frac{0.25}{1} = 0.25:0.25\] The available amounts of \(\mathrm{H}_2\) and \(\mathrm{O}_2\) are equal to the stoichiometric ratio, meaning that in this case there is no limiting reagent. g. For \( 5.00 \ \mathrm{g} \ \mathrm{H}_{2} \) and \( 56.00 \ \mathrm{g} \ \mathrm{O}_{2} \): First, convert the grams of \( \mathrm{H}_{2} \) and \( \mathrm{O}_{2} \) to moles (molar mass of \( \mathrm{O}_{2} = 32 \ \mathrm{g/mol} \)): \[ \frac{5.00 \ \mathrm{g} \ \mathrm{H}_{2}}{2 \ \mathrm{g/mol}} = 2.5 \ \mathrm{mol} \ \mathrm{H}_{2}\] \[ \frac{56.00 \ \mathrm{g} \ \mathrm{O}_{2}}{32 \ \mathrm{g/mol}} = 1.75 \ \mathrm{mol} \ \mathrm{O}_{2} \] Calculate the ratio of the moles: \[\frac{2.5}{2}: \frac{1.75}{1} = 1.25:1.75\] Since there are fewer available moles of \(\mathrm{H}_2\) as compared to the stoichiometric ratio, the limiting reagent is \( \mathrm{H}_{2} \).

Key concepts

Stoichiometry

Stoichiometry helps us understand the quantitative relationships in chemical reactions. In the given reaction between hydrogen and oxygen to form water, stoichiometry tells us how much of each reactant is needed or produced. The balanced chemical equation:

• 2 molecules of hydrogen (\( \mathrm{H}_2 \)) and 1 molecule of oxygen (\( \mathrm{O}_2 \)) react to form 2 molecules of water (\( \mathrm{H}_2\mathrm{O} \)).

This means that the stoichiometry ratio is \( 2:1:2 \). By analyzing this ratio, we can determine how much reactant is needed for a complete reaction or how much product is formed. Understanding these ratios allows us to optimize the consumption of reactants or predict how much of a product can be manufactured. The concept of stoichiometry is fundamental in determining limiting reagents, where we explore which reactant gets used up first during the chemical reaction.

Mole Concept

The mole concept is a central component in chemistry that relates the number of particles, like atoms or molecules, to a measurable amount of substance. One mole of a substance contains Avogadro’s number of entities, which is about \( 6.022 \times 10^{23} \). This concept is vital when comparing the amounts of reactants and products in a chemical reaction. By converting weight or volume of a substance into moles, chemists can work with tangible numbers, especially with gases, where gases at STP occupy 22.4 liters per mole.

• In practice, the mole allows us to switch between mass, volume, and number of particles.
• This conversion is critical for solving chemical equations and understanding the scale of reactions.

For the reaction between \( \mathrm{H}_2 \) and \( \mathrm{O}_2 \), understanding moles helps us predict how much water (\( \mathrm{H}_2 \mathrm{O} \)) would be produced from given quantities of hydrogen and oxygen. The mole concept thus directly supports stoichiometric calculations, helping pinpoint the limiting reagent.

Chemical Reaction Balancing

Chemical reaction balancing ensures that the same number of each type of atom appears on both sides of the equation. This is essential because atoms are neither created nor destroyed in a chemical reaction.

• In our reaction, \( 2 \mathrm{H}_2(g) + \mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2\mathrm{O}(g) \), balancing is crucial for preserving the mass and achieving stoichiometry.
• Balancing factors in, determining the precise proportions of hydrogen to oxygen.

By ensuring correct stoichiometric coefficients, balancing the equation helps to accurately determine which reagent will be limiting. It's fundamental in shaping how much of each compound participates in the reaction, and it forms the framework for calculations concerning moles and grams.

Molar Mass Calculations

Molar mass calculations connect mass to moles, acting as the bridge for converting a substance's mass to its quantity in moles. This involves summing the atomic masses from the periodic table for a compound.

• The molar mass of \( \mathrm{H}_2 \) is \( 2 \mathrm{g/mol} \), while for \( \mathrm{O}_2 \), it is \( 32 \mathrm{g/mol} \).

This calculation is imperative in evaluating how much of each reactant is available when finding the limiting reagent. The calculation is achieved using the formula:

• \( \text{Moles of substance} = \frac{\text{Mass (g)}}{\text{Molar mass (g/mol)}} \)

For the reaction scenarios, molar mass helps transform mass values into a comparable quantity of moles. This step is critical in understanding the complete reaction process and determining which reactant is consumed first.