Question

What volume of \(0.2500 \mathrm{M}\) cobalt(III) sulfate is required to react completely with (a) \(25.00 \mathrm{~mL}\) of \(0.0315 \mathrm{M}\) calcium hydroxide? (b) \(5.00 \mathrm{~g}\) of sodium carbonate? (c) \(12.50 \mathrm{~mL}\) of \(0.1249 \mathrm{M}\) potassium phosphate?

Short answer

Question: Find the volume of 0.2500 M Co(SO₄)₃ required to react completely with the given amounts and concentrations of other compounds: (a) 0.0315 M Ca(OH)₂ (25.00 mL), (b) 5.00 g of Na₂CO₃, and (c) 0.1249 M K₃PO₄ (12.50 mL). Answer: (a) 0.525 mL of cobalt(III) sulfate is required. (b) 62.92 mL of cobalt(III) sulfate is required. (c) 4.16 mL of cobalt(III) sulfate is required.

Step-by-step solution

Balanced equation

Co(SO₄)₃ + 6Ca(OH)₂ → 2Ca₃(PO₄)₂ + 3Co(OH)₂ + 6H₂O

Moles of Ca(OH)₂

The moles of Ca(OH)₂ are given by the product of volume and molarity: moles(Ca(OH)₂) = \(0.0315 \mathrm{M} \times 0.02500 \mathrm{L} = 0.0007875 \mathrm{mol}\).

Stoichiometry

From the balanced equation, we see that 1 mole of Co(SO₄)₃ reacts with 6 moles of Ca(OH)₂. So, moles(Co(SO₄)₃) = (1/6) * moles(Ca(OH)₂) = (1/6) * 0.0007875 = 0.00013125 mol.

Volume of Co(SO₄)₃

Now, we can find the volume of Co(SO₄)₃ required, using the formula: volume(L) = moles / molarity = 0.00013125 / 0.2500 = 0.000525 L. Converting to milliliters, we have: volume = \(0.525 \mathrm{mL}\). Therefore, \(0.525 \mathrm{mL}\) of cobalt(III) sulfate is required. (b) Reaction with sodium carbonate: We first need to find the balanced chemical equation and molarity when reacting with sodium carbonate (Na₂CO₃). The balanced equation is given by:

Balanced equation

Co(SO₄)₃ + 3Na₂CO₃ → Co₂(CO₃)₃ + 3Na₂SO₄

Moles of Na₂CO₃

To find the moles of Na₂CO₃, we need to calculate the moles in 5.00 g: moles(Na₂CO₃) = mass / molar_mass, where the molar mass of Na₂CO₃ is 105.988 g/mol. Moles(Na₂CO₃) = 5.00 / 105.988 = 0.0472 mol.

Stoichiometry

From the balanced equation, we see that 1 mole of Co(SO₄)₃ reacts with 3 moles of Na₂CO₃. So, moles(Co(SO₄)₃) = (1/3) * moles(Na₂CO₃) = (1/3) * 0.0472 = 0.01573 mol.

Volume of Co(SO₄)₃

Now, we need to find the volume of Co(SO₄)₃ required, using the formula: volume(L) = moles / molarity = 0.01573 / 0.2500 = 0.06292 L. Converting to milliliters, we have: volume = \(62.92 \mathrm{mL}\). Therefore, \(62.92 \mathrm{mL}\) of cobalt(III) sulfate is required. (c) Reaction with potassium phosphate: First, let's find the balanced chemical equation and moles for the reaction with potassium phosphate (K₃PO₄). The balanced equation is given by:

Balanced equation

2Co(SO₄)₃ + 3K₃PO₄ → Co₂(PO₄)₃ + 3K₂SO₄

Moles of K₃PO₄

The moles of K₃PO₄ are given by the product of volume and molarity: moles(K₃PO₄) = 0.1249 M * 0.01250 L = 0.001562 mol.

Stoichiometry

From the balanced equation, we see that 2 moles of Co(SO₄)₃ react with 3 moles of K₃PO₄. So, moles(Co(SO₄)₃) = (2/3) * moles(K₃PO₄) = (2/3) * 0.001562 = 0.001041 mol.

Volume of Co(SO₄)₃

Now, we can find the volume of Co(SO₄)₃ required, using the formula: volume(L) = moles / molarity = 0.001041 / 0.2500 = 0.00416 L. Converting to milliliters, we have: volume = \(4.16 \mathrm{mL}\). Therefore, \(4.16 \mathrm{mL}\) of cobalt(III) sulfate is required.

Key concepts

Molar Volume

Molarity and volume are closely related concepts used to describe concentrations in chemistry. Molar volume is the volume occupied by one mole of substance. At standard temperature and pressure (STP), the molar volume of a gas is approximately 22.4 liters per mole. However, molar volume can vary with conditions and states of matter. For liquid solutions like the cobalt(III) sulfate mentioned in the exercise, molar volume isn’t a fixed value like it is for gases at STP but rather a way to describe the volume that a known amount of substance occupies when dissolved to make a certain molarity of solution.

For instance, if you have a 1.0 Molar (1.0 M) solution, this means there is 1 mole of solute in every 1 liter of solution. If the problem requires 0.2500 M cobalt(III) sulfate, we can use the molarity and the amount of solute needed to find the volume needed for the reaction. In the context of solving stoichiometry problems, understanding how to use molar volume is crucial when dealing with gases or converting between moles and liters for solutions.

Chemical Reaction Equations

Chemical reaction equations give us a snapshot of the substances involved in a reaction, both reactants and products, as well as their quantities in terms of moles. A balanced chemical equation has the same number of atoms for each element on both the reactants and products side, complying with the Law of Conservation of Mass. For example, in part (a) of the exercise, the balanced equation Co(SO₄)₃ + 6Ca(OH)₂ → 2Ca₃(PO₄)₂ + 3Co(OH)₂ + 6H₂O shows that one mole of cobalt(III) sulfate reacts with six moles of calcium hydroxide to produce other compounds.

Understanding how to write and balance chemical equations is fundamental in stoichiometry as it helps you to visualize and convert between moles of different reactants and products, utilizing the stoichiometric coefficients (numbers in front of molecules) which serve as mole-to-mole ratios.

Molarity Calculations

Molarity, often denoted by M, is a measure of concentration that tells us how many moles of a substance are present in one liter of solution. It is calculated by dividing the number of moles of solute by the volume of solution in liters. Molarity calculations are central to problems where solutions' concentrations and volumes are considered, such as the one presented in the exercise. For example, in the step-by-step solution, we use molarity to determine the moles of reactants and thus can calculate the volume of cobalt(III) sulfate required for the reaction.

Understanding how to perform molarity calculations allows for the conversion between the volume of a solution and the amount of substance it contains, which is a critical step in solving stoichiometry problems involving solutions.

Mole-to-Mole Ratios

Mole-to-mole ratios are derived from the coefficients of compounds in a balanced chemical equation. They tell us how many moles of one substance will react or form from a certain number of moles of another substance. In the exercise, these ratios are crucial in determining how much of one compound will react with a given amount of another. For instance, the balanced equation for part (a) indicates that for every mole of cobalt(III) sulfate, six moles of calcium hydroxide are needed. By using the mole-to-mole ratio, we can convert between moles of different chemicals involved in the reaction.

Understanding and applying these ratios is an essential skill in stoichiometry. It enables students to progress from knowing the amount of a reactant to finding the required amount of another reactant or product necessary for a complete reaction, ensuring that problems like the ones in the exercise are solved accurately.