Question

Find the mass of barium metal (in grams) that must react with \(\mathrm{O}_{2}\) to produce enough barium oxide to prepare \(1.0 \mathrm{~L}\) of a \(0.10 \mathrm{M}\) solution of \(\mathrm{OH}^{-}\). (Hint: Barium metal reacts with oxygen to form \(\mathrm{BaO} ; \mathrm{BaO}\) reacts with water to form \(\left.\mathrm{Ba}(\mathrm{OH})_{2} .\right)\)

Short answer

To prepare a 1.0 L solution of 0.10 M OH⁻, the mass of barium (Ba) required is the product of the moles of Ba and the molar mass of Ba, calculated stepwise from the moles of OH⁻.

Step-by-step solution

Calculate moles of Hydroxide (OH⁻)

Determine the number of moles of hydroxide ions (OH⁻) needed. Use the molarity (M) of the OH⁻ solution and the volume (L) to calculate the moles. The molarity is given as 0.10 M and the volume of the solution is 1.0 L. The number of moles of OH⁻ (n) is equal to molarity (M) multiplied by volume (V): n(OH⁻) = M × V = 0.10 M × 1.0 L.

Calculate moles of Barium Hydroxide (Ba(OH)₂)

Since one molecule of barium hydroxide (Ba(OH)₂) dissociates into two hydroxide ions (OH⁻), the number of moles of Ba(OH)₂ required is half of the number of moles of OH⁻. n(Ba(OH)₂) = n(OH⁻) / 2.

Convert moles of Barium Hydroxide to moles of Barium Oxide (BaO)

Barium oxide reacts with water to form barium hydroxide: BaO + H₂O → Ba(OH)₂. This reaction shows a 1:1 mole ratio between BaO and Ba(OH)₂. Therefore, the number of moles of BaO required is equal to the number of moles of Ba(OH)₂ calculated in Step 2.

Calculate moles of Barium (Ba)

Barium reacts with oxygen (O₂) to form barium oxide (BaO): 2 Ba + O₂ → 2 BaO. Since we have two moles of Ba for every mole of BaO produced, we need half the number of moles of BaO to determine the moles of Ba: n(Ba) = n(BaO) / 2.

Calculate mass of Barium (Ba)

Using the molar mass of barium, convert the moles of barium calculated in Step 4 to grams. The molar mass of barium (Ba) is approximately 137.33 g/mol. Mass (Ba) = n(Ba) × molar mass of Ba.

Key concepts

Chemical Reaction Equations

Understanding the math behind chemistry involves interpreting chemical reaction equations. These equations are like recipes for how substances, or reactants, transform into different substances, known as products. In the exercise, the reaction of barium metal with oxygen to form barium oxide, followed by its reaction with water to produce barium hydroxide, is considered.

For a clearer grasp of the process, the reaction equation for barium metal with oxygen is written as:
\[2 \mathrm{Ba} + \mathrm{O}_2 \rightarrow 2 \mathrm{BaO}\]
This equation indicates that two atoms of barium (Ba) react with one molecule of oxygen (\(\mathrm{O}_2\)) to produce two units of barium oxide (BaO). Reading these equations effectively allows us to apply stoichiometry, which is the quantitative relationship between reactants and products in a chemical reaction.

Mole Concept

The mole concept is a bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure. One mole is Avogadro's number (approximately \(6.022 \times 10^{23}\)) of particles, whether they are atoms, molecules, ions, or electrons. In the given exercise, we work with moles to figure out how much barium metal we need.

We used the formula:
\[ n(\mathrm{OH}^-) = M \times V \]
Where \(n\) represents the number of moles, \(M\) is the molarity, and \(V\) is the volume in liters. By strategically using moles, we connect the concentration of the hydroxide ion solution (OH⁻) with the amount of barium oxide and ultimately barium metal needed. It's a stepping stone to converting abstract numbers into tangible amounts of substances we can measure.

Molar Mass

The molar mass is the weight of one mole of a substance, typically expressed in grams per mole (g/mol). It acts as a conversion factor between the mass of a sample and the number of moles contained in that sample. The atomic or molecular weight of an element or compound, usually found on the periodic table or by summing the weights of the constituent elements, gives us this crucial value.

In our problem, we calculate the mass of barium needed using its molar mass (137.33 g/mol) and the moles of barium derived from stoichiometric calculations. This is shown in the last step, utilizing the formula:
\[ \text{Mass (Ba)} = n(\text{Ba}) \times \text{molar mass of Ba} \]
By multiplying the number of moles by the molar mass, we can transition from an abstract concept (moles) to a more practical one (grams) that can be physically measured on a scale, closing the gap between theory and practice in chemistry.