Question

The equation for the combustion of hydrogen is $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) What is the mole ratio of \(\mathrm{O}_{2}\) to \(\mathrm{H}_{2}\) ? (b) If \(1.0 \mathrm{~g}\) of \(\mathrm{H}_{2}\) reacts, what mass of \(\mathrm{O}_{2}\) will react with it, and what mass of \(\mathrm{H}_{2} \mathrm{O}\) should form? (c) When \(1.0 \mathrm{~g} \mathrm{H}_{2}\) is mixed with \(4.0 \mathrm{~g} \mathrm{O}_{2}\), what is the theoretical yield of \(\mathrm{H}_{2} \mathrm{O}\) ?

Short answer

The mole ratio of \(O_{2}\) to \(H_{2}\) is 1:2. When 1.0 g of \(H_{2}\) reacts, approximately 8 g of \(O_{2}\) react with it, and approximately 18 g of \(H_{2}O\) form. When 1.0 g of \(H_{2}\) is mixed with 4.0 g of \(O_{2}\), the theoretical yield of \(H_{2}O\) is approximately 4.5 g.

Step-by-step solution

Mole Ratio Calculation

The balanced chemical equation is: \(2 H_{2}(g) + O_{2}(g) \rightarrow 2 H_{2}O(g)\). This shows that 2 moles of hydrogen (\(H_{2}\)) react with 1 mole of oxygen (\(O_{2}\)) to produce 2 moles of water (\(H_{2}O\)). Therefore, the mole ratio of \(O_{2}\) to \(H_{2}\) is 1:2.

Mass of \(O_{2}\) and \(H_{2}O\) Calculation

Starting with 1.0 g of \(H_{2}\), first convert it into moles. As the molar mass of \(H_{2}\) is approximately 2 g/mol, there is approximately 0.5 moles of \(H_{2}\). If this reacts completely, 0.25 moles of \(O_{2}\) (half the amount as the mole ratio is 1:2) is needed. Considering the molar mass of \(O_{2}\) is approximately 32 g/mol, this equates to approximately 8 g of \(O_{2}\). Approximately 1 mole of \(H_{2}O\) should form, which equates to an approximate mass of 18 g.

Theoretical Yield Calculation

If 1.0 g of \(H_{2}\) is mixed with 4.0 g of \(O_{2}\), this equates to approximately 0.5 moles of \(H_{2}\) and 0.125 moles of \(O_{2}\). As per the stoichiometry of the reaction, 0.25 moles of \(O_{2}\) are needed for 0.5 moles of \(H_{2}\). Here, \(O_{2}\) is the limiting reactant because it is not present in enough quantity to react with all the \(H_{2}\). Therefore, the theoretical yield is based on the moles of \(O_{2}\). This equates to a yield of approximately 0.25 moles of \(H_{2}O\), or approximately 4.5 g.

Key concepts

Understanding Mole Ratio

When discussing chemical reactions, one key concept to understand is the mole ratio. Imagine you're following a recipe. Just like you need a certain ratio of ingredients to bake a cake, chemical reactions need specific ratios of chemicals to react properly. In our chemical equation for the combustion of hydrogen, we see:

• 2 \(\mathrm{H}_{2}(g)\)reacts with 1 \(\mathrm{O}_{2}(g)\)
• to form 2 \(\mathrm{H}_{2}\mathrm{O}(g)\).

This gives us the mole ratio of \(\mathrm{O}_{2}\) to \(\mathrm{H}_{2}\), which is 1:2.
1 mole of oxygen is required to react with 2 moles of hydrogen. This ratio is crucial for calculating other values like the amount of reactants needed or the products formed.
When you solve a chemistry problem, always look at the balanced equation to find the correct mole ratios.
Mole ratios act as a bridge when converting between moles of different substances in a chemical equation. They help ensure that everything is proportionate and chemically correct during the calculations.

Identifying the Limiting Reactant

Every chemical reaction involves reactants that interact to produce products. However, one reactant will usually run out before the others, halting the reaction. This is known as the limiting reactant. In our scenario:

• 1.0 g of \(\mathrm{H}_{2}\)provides around 0.5 moles.
• While 4.0 g of\(\mathrm{O}_{2}\)
• provides only around 0.125 moles.

Given the reaction requires 0.25 moles of \(\mathrm{O}_{2}\) for 0.5 moles of \(\mathrm{H}_{2}\), we see that oxygen is in short supply. Hence, \(\mathrm{O}_{2}\)is the limiting reactant here.
Identifying the limiting reactant is essential as it dictates the maximum amount of product that can be formed in the reaction. In essence:
- Find the moles of each reactant available.- Compare their ratios using the balanced equation.- Identify which reactant runs out first.
Understanding this concept ensures the calculations are based on the most accurate chemical scenario.

Calculating Theoretical Yield

Theoretical yield is the maximum amount of product that can be formed from a given amount of reactants in a chemical reaction.
Let's break it down:

• Here, we use the limiting reactant to determine the theoretical yield.
• In the case of the combustion of hydrogen:\(\mathrm{O}_{2}\)is the limiting factor

This means the calculation for potential product will be based on it.
From 0.125 moles of \(\mathrm{O}_{2}\), the reaction produces 0.25 moles of\(\mathrm{H}_{2} \mathrm{O}\).
The molar mass of \(\mathrm{H}_{2} \mathrm{O}\)is approximately 18 g/mol. Thus, the theoretical yield is:\(0.25 \times 18 = 4.5\) g.A practical rule to remember is:- Always identify and base the theoretical yield calculations on the limiting reactant.- The theoretical yield helps chemists understand the efficiency of a reaction by comparing it against actual yields obtained during experiments.