Question

Consider the following unbalanced equation: \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{CaSO}_{4}(s)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q)\) What masses of calcium sulfate and phosphoric acid can be produced from the reaction of \(1.0 \mathrm{~kg}\) calcium phosphate with \(1.0\) \(\mathrm{kg}\) concentrated sulfuric acid \(\left(98 \% \mathrm{H}_{2} \mathrm{SO}_{4}\right.\) by \(\left.\mathrm{mass}\right)\) ?

Short answer

The masses of calcium sulfate and phosphoric acid that can be produced from the reaction of 1.0 kg calcium phosphate with 1.0 kg of 98% sulfuric acid are \(1.316\, \text{kg}\) and \(0.6315\, \text{kg}\), respectively.

Step-by-step solution

Balance the chemical equation

To balance the given chemical equation, determine the appropriate stoichiometric coefficients for each compound. The balanced equation is: \(\mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2}(s) + 3\mathrm{H}_{2}\mathrm{SO}_{4}(a q) \longrightarrow 3\mathrm{CaSO}_{4}(s)+ 2\mathrm{H}_{3}\mathrm{PO}_{4}(a q)\)

Calculate the moles of reactants

Calculate the moles of calcium phosphate and sulfuric acid given their masses: - Calcium phosphate: \(1.0\,\text{kg} = 1000\,\text{g}\). The molar mass of \(\mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2}\) is \(3 \times 40.08 + 2 \times (1 \times 30.97 + 4 \times 16.00) = 310.18\, \text{g/mol}\) The moles of calcium phosphate = \(\frac{1000\, \text{g}}{310.18\, \text{g/mol}} = 3.223\, \text{mol}\) - Concentrated sulfuric acid: 98% by mass means that for each kilogram, there are 0.98 kg of \(\mathrm{H}_{2}\mathrm{SO}_{4}\). The molar mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) is \(2 \times 1.01 + 32.06 + 4 \times 16.00 = 98.08\, \text{g/mol}\). The moles of sulfuric acid = \(\frac{0.98 \times 1000\, \text{g}}{98.08\, \text{g/mol}} = 9.986\, \text{mol}\)

Identify the limiting reactant

Using the balanced equation, the mole ratio of calcium phosphate to sulfuric acid is 1:3, so for every mole of calcium phosphate, we need 3 moles of sulfuric acid. Divide the moles of sulfuric acid by the moles of calcium phosphate and compare the resulting ratio to the stoichiometric ratio to determine the limiting reactant: \(\frac{9.986\,\text{mol}}{3.223\,\text{mol}} = 3.10\) This ratio is slightly larger than 3, it means we have a small excess of sulfuric acid, so calcium phosphate is the limiting reactant.

Determine the masses of the products

From the balanced equation, we find the following stoichiometric relationships between reactants and products: (calcium phosphate): (calcium sulfate) : (phosphoric acid) = 1:3:2 Using the moles of limiting reactant (calcium phosphate): - Moles of calcium sulfate produced: \(3.223\,\text{mol}\times 3 = 9.669\, \text{mol}\) The molar mass of \(\mathrm{CaSO}_4\) is \(40.08 + 32.06 + 4 \times 16.00 = 136.14\, \text{g/mol}\). Mass of calcium sulfate = \(9.669\, \text{mol} \times 136.14\, \text{g/mol} = 1316.31\, \text{g} = 1.316\, \text{kg}\) - Moles of phosphoric acid produced: \(3.223\, \text{mol} \times 2 = 6.446\, \text{mol}\) The molar mass of \(\mathrm{H}_{3}\mathrm{PO}_{4}\) is \(3 \times 1.01 + 30.97 + 4 \times 16.00 = 97.99\, \text{g/mol}\). Mass of phosphoric acid = \(6.446\, \text{mol} \times 97.99\, \text{g/mol} = 631.55\, \text{g}\) So, the masses of calcium sulfate and phosphoric acid that can be produced from the reaction of 1.0 kg calcium phosphate with 1.0 kg of 98% sulfuric acid are 1.316 kg and 0.6315 kg, respectively.

Key concepts

Limiting Reactant

Understanding the concept of the limiting reactant is critical in stoichiometry, which is the study of the amounts of substances consumed and produced in chemical reactions. Imagine baking cookies: if you can make 10 cookies with 1 egg but only have enough flour for 5, then flour is your limiting ingredient, just as the limiting reactant in a chemical reaction determines the amount of product formed.

In the given exercise, we used the stoichiometric coefficients from the balanced chemical equation to compare the mole ratio of reactants. By comparing the actual mole ratio to the stoichiometric ratio, we identified the limiting reactant. In our case, calcium phosphate limits the reaction, meaning it will be consumed first, and no more products can be formed once it is used up, just as no more cookies can be baked without additional flour.

Mole Concept

The mole concept is a bridge between the microscopic world of atoms and molecules and the macroscopic world we live in. Just as a dozen refers to a count of 12 of any objects, a mole represents approximately 6.022 x 10^23 entities, be they atoms or molecules—a number known as Avogadro's constant. The beauty of using moles in chemistry lies in simplifying the comparison of amounts of different substances based on a common scale.

In the exercise, the mole concept allowed us to convert the given masses of calcium phosphate and sulfuric acid into amounts that can be directly compared using their respective molar masses. This made it possible to calculate the moles of each reactant and subsequently determine the limiting reactant—essential steps for predicting the amount of product that can be produced in a chemical reaction.

Molar Mass

Molar mass is the weight of one mole of a substance, typically expressed in grams per mole (g/mol), and it's the molecular weight or atomic weight reported on the periodic table for each element. It's like knowing the weight of a dozen eggs; if one egg weighs 50 grams, a dozen would weigh 600 grams. Molar mass lets us convert between the weight of a substance in grams and the number of moles simply and effectively.

We utilized the molar mass in our exercise to convert grams of reactants into moles for stoichiometric calculations. Knowing that molar mass is a unique characteristic of each compound, we calculated it based on the atomic weights of the elements that make up calcium phosphate and sulfuric acid, enabling us to find the moles of each compound from the masses given and proceed with determining the limiting reactant and predicting the mass of products formed.