Question

If \(10.0 \mathrm{g}\) of hydrogen gas is reacted with \(10.0 \mathrm{g}\) of oxygen gas according to the equation \\[ 2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \\] we should not expect to form \(20.0 \mathrm{g}\) of water. Why not? What mass of water can be produced with a complete reaction?

Short answer

We should not expect to form 20.0 grams of water because the reaction depends on the amount of the limiting reactant, oxygen, which is less than what is required for a complete reaction with 10.0 grams of hydrogen gas. The mass of water produced in this reaction is 11.26 grams.

Step-by-step solution

Write down the given information

We are given that 10.0 g of hydrogen gas (H₂) and 10.0 g of oxygen gas (O₂) are reacted, according to the equation: \[ 2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\ \]

Calculate the moles of reactants

To determine the limiting reactant, first we need to convert grams of reactants to moles. We'll use the molar masses. For hydrogen gas (H₂): \\(M_{H2} = 2.016 \frac{g}{mol}\\), Oxygen gas (O₂): \\(M_{O2} = 32.00 \frac{g}{mol}\\) \[ Moles \ of \ H_{2} = \frac{10.0 \ \mathrm{g}}{2.016 \ \frac{g}{mol}} = 4.960 \ \mathrm{mol} \ \mathrm{H_{2}} \] \[ Moles \ of \ O_{2} = \frac{10.0 \ \mathrm{g}}{32.00 \ \frac{g}{mol}} = 0.3125 \ \mathrm{mol} \ \mathrm{O_{2}} \]

Determine the limiting reactant

To find the limiting reactant, we need to compare the ratio of the moles of reactants to their stoichiometric coefficients in the balanced equation. The balanced equation can be converted to the ratio of molar amounts: \[ \frac{2 \ \mathrm{mol} \ \mathrm{H_{2}}}{1 \ \mathrm{mol} \ \mathrm{O_{2}}} = \frac{4.960 \ \mathrm{mol} \ \mathrm{H_{2}}}{\mathrm{x \ mol \ O_{2}}} \] Solving for x in the above equation: \[ x = \frac{4.960 \ \mathrm{mol} \ \mathrm{H_{2}}}{2} = 2.480 \ \mathrm{mol} \ \mathrm{O_{₂}} \] Since we only have 0.3125 mol of oxygen, oxygen is the limiting reactant.

Use stoichiometry to find moles of water produced

Now we'll use stoichiometry and the limiting reactant, oxygen, to find moles of water produced: From the balanced chemical equation: \[ 1 \ \mathrm{mol} \ \mathrm{O_{2}} \rightarrow 2 \ \mathrm{mol} \ \mathrm{H_{2}O} \] So, as we have 0.3125 mol of oxygen gas, \[ 0.3125 \ \mathrm{mol} \ \mathrm{O_{2}} \rightarrow 2 \times 0.3125 \ \mathrm{mol} \ \mathrm{H_{2}O} = 0.625 \ \mathrm{mol} \ \mathrm{H_{2}O} \]

Calculate the mass of water produced

Finally, we'll convert the moles of water produced into grams. The molar mass of water (H₂O) is 18.02 g/mol. \[ Mass \ of \ H_{2}O = 0.625 \ \mathrm{mol} \ \mathrm{H_{2}O} \times 18.02 \ \frac{g}{mol} = 11.26 \ \mathrm{g} \] So, the mass of water produced in this reaction is 11.26 grams. The reason we should not expect to form 20.0 grams of water is because the reaction between hydrogen and oxygen gas is not solely governed by their initial mass. Instead, it depends on the amount of oxygen available, i.e., the limiting reactant.

Key concepts

Stoichiometry

Stoichiometry is the field of chemistry that is concerned with the relative quantities of reactants and products in chemical reactions. It is based on the conservation of mass where the total mass of the reactants equals the total mass of the products. By understanding stoichiometry, we can predict the amounts of substances consumed and produced in a reaction.

For instance, in the reaction between hydrogen and oxygen to produce water, stoichiometry tells us that two molecules of hydrogen gas react with one molecule of oxygen gas to produce two molecules of water. This is represented in the balanced chemical equation: \(2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O}\). To apply stoichiometry to an exercise, we convert the given masses of the reactants to moles, because reactions occur at the molecular level, between atoms and molecules, and not between grams of substances.

Molar Mass

Understanding Molar Mass

The molar mass is a fundamental concept in stoichiometry that represents the mass of one mole of a substance. It is numerically equal to the average mass of the molecule expressed in atomic mass units (amu) and provides a bridge between the mass of a substance and the number of particles it contains.

In the given exercise, to find the amount of substance in moles, we apply the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). For example, hydrogen has a molar mass of \(2.016 \frac{g}{mol}\) and oxygen has a molar mass of \(32.00 \frac{g}{mol}\). These values allow us to calculate the moles of each reactant in the chemical reaction which is the initial step in identifying the limiting reactant.

Balanced Chemical Equations

The Role of Balanced Equations in Reactions

Balanced chemical equations are representations of chemical reactions that obey the law of conservation of mass. The number of atoms of each element on the reactant side must be equal to the number of atoms of that element on the product side. By balancing the equation, we can use it to understand the mole ratio of reactants and products.

In our water formation exercise, the balanced equation \(2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O}\) tells us that two moles of hydrogen gas react with one mole of oxygen gas to produce two moles of water. This ratio is crucial when determining the amounts of reactants needed and the amount of product expected - it is the stoichiometric ratio.

Chemical Reaction Yield

Actual vs. Theoretical Yield

Chemical reaction yield refers to the amount of product obtained from a chemical reaction. In theory, the yield would be equal to the amount predicted by stoichiometry, known as the theoretical yield. However, in reality, due to various factors such as incomplete reactions or side reactions, the actual yield is often less.

The concept of limiting reactant is essential for understanding yield. The limiting reactant is the substance that is totally consumed first and thus determines the amount of product that can be formed. In our exercise, the oxygen gas is the limiting reactant, which imposes a limit on the amount of water that can be produced - leading to the actual yield being just 11.26 grams instead of 20.0 grams as one might simplistically expect if not considering the stoichiometric ratios and limiting reactants.