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 \(\left(98 \% \mathrm{H}_{2} \mathrm{SO}_{4}\text { by mass)? }\right.\)

Short answer

In summary, when reacting 1.0 kg of calcium phosphate with 1.0 kg of concentrated sulfuric acid (98% H₂SO₄ by mass), 1314.71 g of calcium sulfate and 630.62 g of phosphoric acid will be produced.

Step-by-step solution

Balance the chemical equation

We need to balance the chemical equation given: $$\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{H}_{2}\mathrm{SO}_{4}(a q) \longrightarrow \mathrm{CaSO}_{4}(s)+\mathrm{H}_{3}\mathrm{PO}_{4}(a q)$$ The balanced equation is: $$\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) + 3\mathrm{H}_{2}\mathrm{SO}_{4}(a q) \longrightarrow 3\mathrm{CaSO}_{4}(s) + 2\mathrm{H}_{3}\mathrm{PO}_{4}(a q)$$

Calculate the moles of reactants

Next, we need to calculate the moles of calcium phosphate (Ca3(PO4)2) and sulfuric acid (H2SO4). We are given 1.0 kg of calcium phosphate and 1.0 kg of 98% concentrated sulfuric acid. First, let's convert the masses to moles. The molar mass of calcium phosphate Ca3(PO4)2 is: \(3 \times \mathrm{Ca} + 2\times (1\times \mathrm{P} + 4\times \mathrm{O}) = (3\times40.08)+(2\times (1\times30.97 + 4\times15.999)) = 310.18 \: g/mol\) The molar mass of sulfuric acid H2SO4 is: \(2\times \mathrm{H} + 1\times \mathrm{S} + 4\times \mathrm{O} = (2\times1.01)+(1\times32.07)+(4\times15.999) = 98.08 \: g/mol\) Now, we calculate the moles: Moles of calcium phosphate = mass / molar mass, which is \(1.0\mathrm{kg} \times \dfrac{1000\mathrm{g}}{1\mathrm{kg}} \times \dfrac{1 \: mol}{310.18 \: g}= \dfrac{1000}{310.18} \: mol = 3.22 \: mol\) We have 98% concentrated sulfuric acid, so the mass of pure sulfuric acid is: \(1.0\mathrm{kg} \times 0.98 = 0.98\mathrm{kg}\) Moles of sulfuric acid = mass / molar mass, which is \(0.98\mathrm{kg} \times \dfrac{1000\mathrm{g}}{1\mathrm{kg}} \times \dfrac{1 \: mol}{98.08 \: g}= \dfrac{980}{98.08} \: mol = 9.99 \: mol\)

Identify the limiting reactant

Now, we need to identify the limiting reactant. We will compare the mole ratio of calcium phosphate and sulfuric acid to the ratio in the balanced equation. Mole ratio of reactants = 3.22 mol Ca3(PO4)2 / 9.99 mol H2SO4 = 0.322 Mole ratio in the balanced equation = 1 mol Ca3(PO4)2 / 3 mol H2SO4 = 0.333 Since the mole ratio of reactants is less than the ratio in the balanced equation, calcium phosphate is the limiting reactant.

Calculate the moles of products

Using the limiting reactant (calcium phosphate) and the balanced chemical equation, we can calculate the moles of calcium sulfate (CaSO4) and phosphoric acid (H3PO4). From the balanced equation, 1 mol of calcium phosphate reacts with 3 mol of sulfuric acid to produce 3 mol of calcium sulfate and 2 mol of phosphoric acid. Moles of calcium sulfate produced = 3.22 mol Ca3(PO4)2 x (3 mol CaSO4 / 1 mol Ca3(PO4)2) = 3.22 x 3 = 9.66 mol Moles of phosphoric acid produced = 3.22 mol Ca3(PO4)2 x (2 mol H3PO4 / 1 mol Ca3(PO4)2) = 3.22 x 2 = 6.44 mol

Convert moles to mass

Finally, we can convert the moles of calcium sulfate and phosphoric acid to mass. Molar mass of calcium sulfate CaSO4 is: \(1\times \mathrm{Ca} + 1\times \mathrm{S} + 4\times \mathrm{O} = (1\times40.08)+(1\times32.07)+(4\times15.999) = 136.14 \: g/mol\) Molar mass of phosphoric acid H3PO4 is: \(3\times \mathrm{H} + 1\times \mathrm{P} + 4\times \mathrm{O} =(3\times1.01)+(1\times30.97)+(4\times15.999) = 97.99 \: g/mol\) Mass of calcium sulfate = 9.66 mol x 136.14 g/mol = 1314.71 g Mass of phosphoric acid = 6.44 mol x 97.99 g/mol = 630.62 g The masses of calcium sulfate and phosphoric acid produced are 1314.71 g and 630.62 g, respectively.

Key concepts

Limiting Reactant

When we have a chemical reaction, the limiting reactant is the one that gets used up first. This reactant determines how much of the products will be formed. Think of it like ingredients for a cake. If you run out of eggs, you can't make more cakes, even if you have a lot of flour and sugar left. In this exercise, we compared the mole ratio of calcium phosphate to sulfuric acid with their ratio in the balanced equation. Since calcium phosphate had fewer moles relative to the requirement, it was the limiting reactant. This means once calcium phosphate is used up, no more products can be formed.

Balancing Chemical Equations

Balancing chemical equations ensures the same number of each type of atom on both sides of the equation. This is based on the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. For the given reaction, we balanced the equation by ensuring there are equal numbers of calcium, phosphorus, oxygen and sulfur atoms on both sides. It was done by adjusting coefficients like having 3 moles of sulfuric acid for every 1 mole of calcium phosphate, ensuring all atoms balance perfectly. This balanced equation is essential for accurately finding mole ratios used in stoichiometry calculations.

Mole Concept

The mole is a fundamental concept in chemistry that serves as a bridge between the atomic world and the macroscopic world. It allows chemists to count atoms, molecules, or ions in a feasible way. One mole contains Avogadro's number of particles, which is approximately \(6.022 \times 10^{23}\). In our problem, we calculated the number of moles of each reactant to eventually determine how much of each product can be made. For example, the number of moles of calcium phosphate and sulfuric acid were computed by dividing the mass of each reactant by its respective molar mass.

Molar Mass Calculation

Calculating molar mass involves adding up the atomic masses of all atoms in a molecule to determine the mass of one mole. This is crucial in converting a substance's mass to moles, allowing for stoichiometric calculations. For example, the molar mass of calcium phosphate \( ext{Ca}_3( ext{PO}_4)_2\) was determined by totaling the mass of calcium, phosphorus, and oxygen atoms. Similarly, we derived the molar mass for sulfuric acid \( ext{H}_2 ext{SO}_4\). These calculations facilitated the conversion of the given masses of reactants to moles, a necessary step in finding out the amounts of products formed.