Question

A hydrogen-filled balloon is ignited and 1.50 \(\mathrm{g}\) of hydrogen reacts with 12.0 \(\mathrm{g}\) of oxygen. How many grams of water vapor form? (Assume that water vapor is the only product.)

Short answer

13.52 grams of water vapor are formed.

Step-by-step solution

Write the balanced chemical equation

The chemical reaction for the combustion of hydrogen is given by: 2 H\(_2\) + O\(_2\) → 2 H\(_2\)O.

Convert the masses of reactants to moles

Calculate the number of moles of hydrogen and oxygen using their respective molar masses. For hydrogen: Molar mass of H\(_2\) is 2.02 g/mol.Moles of H\(_2\) = mass (g) / molar mass (g/mol) = 1.50 g / 2.02 g/mol = 0.743 moles.For oxygen: Molar mass of O\(_2\) is 32.00 g/mol.Moles of O\(_2\) = mass (g) / molar mass (g/mol) = 12.00 g / 32.00 g/mol = 0.375 moles.

Determine the limiting reactant

From the balanced equation, 1 mole of O\(_2\) reacts with 2 moles of H\(_2\). We have 0.743 moles of H\(_2\) and 0.375 moles of O\(_2\). The ratio of available H\(_2\) to required H\(_2\) is:0.743 / 2 = 0.372 moles of H\(_2\) per mole of O\(_2\).Since we only have 0.375 moles of O\(_2\), and it requires 2 moles of H\(_2\) (which we have more than enough), O\(_2\) is the limiting reactant.

Calculate the amount of water produced

Using the mole ratio from the balanced equation, 1 mole of O\(_2\) produces 2 moles of H\(_2\)O.Thus, 0.375 moles of O\(_2\) will produce 0.375 * 2 = 0.750 moles of H\(_2\)O.

Convert the moles of water to grams

The molar mass of H\(_2\)O is 18.02 g/mol.Mass of H\(_2\)O formed = moles of H\(_2\)O * molar mass of H\(_2\)O = 0.750 moles * 18.02 g/mol = 13.52 g.Therefore, 13.52 grams of water vapor are formed.

Key concepts

Balancing Chemical Equations

Balancing chemical equations is a foundational skill in chemistry, essential to solving stoichiometry problems. An equation represents the reactants transforming into products, and it must reflect the conservation of mass. This means the number of atoms for each element must be equal on both sides of the reaction. In the hydrogen combustion example, the equation 2 H₂ + O₂ → 2 H₂O shows a balanced reaction with an equal number of hydrogen and oxygen atoms before and after the reaction.

In practice, balancing an equation often starts with the most complex molecule. You would adjust the coefficients, which are the numbers in front of each molecule, to ensure all atoms balance out. This step is crucial, as it sets the stage for correctly calculating the reactant amounts and predicting the yield of the reaction. Without a balanced equation, all subsequent calculations would be incorrect.

Mole Concept

The mole concept is a bridge between the microscopic world of atoms and molecules and the macroscopic world we experience. One mole represents Avogadro's number, which is approximately 6.022×10²³ entities, whether they're atoms, molecules, ions, or electrons. When dealing with chemical reactions, we refer to moles to express quantities because counting individual atoms is not feasible.

For example, in our stoichiometry problem, we had to convert the given masses of hydrogen and oxygen into moles using their respective molar masses. The molar mass (in grams per mole) of a substance equals the mass of one mole of its particles, making calculations manageable for reactions involving substantial numbers of atoms or molecules. The use of moles is integral to all stoichiometry calculations as it provides a common unit for quantifying the reactants and products in a chemical reaction.

Limiting Reactant

The concept of the limiting reactant is vital when quantities of reactants are not in the exact proportions necessary for a reaction. The limiting reactant is the one that is totally consumed first, thus preventing further reaction and determining the maximum amount of product that can be formed. It's akin to having a car assembly line with plenty of metal and electronics, but a shortage of tires - you can only manufacture as many cars as you have full sets of tires for.

To identify the limiting reactant, we compare the mole ratio of reactants used in the reaction to what we actually have. As shown in the hydrogen and oxygen reaction, despite the excess of hydrogen, the limited amount of oxygen dictates the final amount of water produced. Understanding which reactant limits the reaction is crucial for predicting the outcome and efficiency of a chemical process.

Molar Mass Calculation

Calculating molar mass, the mass of one mole of a substance, is a fundamental step in stoichiometry that facilitates conversion between mass and moles. It is equivalent to the atomic or molecular weight of a substance expressed in grams. One can find the atomic weights on the periodic table which, when added together in the right proportions, give the molar mass of the compound.

In our solution, we calculated the molar mass of diatomic hydrogen (H₂) as 2.02 g/mol and oxygen (O₂) as 32.00 g/mol. With these values, we could convert the given reactant masses into moles. Molar mass calculation is essential for stoichiometry because it aligns the scale of atoms and molecules with measurable quantities, thus allowing chemists to practically use the balanced chemical equations for predicting the masses of reactants and products involved.