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

Short answer

In this balanced chemical equation, $$2 \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2}(s) + 6 \mathrm{H}_{2}\mathrm{SO}_{4}(aq) \longrightarrow 6 \mathrm{CaSO}_{4}(s) + 4 \mathrm{H}_{3}\mathrm{PO}_{4}(aq)$$, from 1.0 kg of calcium phosphate and 1.0 kg of concentrated sulfuric acid (98% by mass), 1315.87 g of calcium sulfate (CaSO₄) and 631.12 g of phosphoric acid (H₃PO₄) can be produced.

Step-by-step solution

Balance the chemical equation

Balancing the given chemical equation, we get: $$ 2 \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2}(s) + 6 \mathrm{H}_{2}\mathrm{SO}_{4}(aq) \longrightarrow 6 \mathrm{CaSO}_{4}(s) + 4 \mathrm{H}_{3}\mathrm{PO}_{4}(aq) $$

Convert the given masses into moles

We need the molar masses of the substances involved: - Ca₃(PO₄)₂: Molar mass = 3(40.08) + 2(1.0079)(4) + 2(30.97)(2) ≈ 310.18 g/mol - H₂SO₄: Molar mass = 2(1.0079) + 32.07 + 4(16.00) ≈ 98.08 g/mol Now we can convert the given masses of 1.0 kg each into moles: - For Ca₃(PO₄)₂: \( \frac{1000 \text{ g}}{310.18 \text{ g/mol}} ≈ 3.22 \text{ mol} \) of Ca₃(PO₄)₂ - For H₂SO₄: since it's 98% by mass, the mass of H₂SO₄ in 1.0 kg of concentrated solution is 0.98 × 1000 g = 980 g. So, \( \frac{980 \text{ g}}{98.08 \text{ g/mol}} ≈ 9.99 \text{ mol} \) of H₂SO₄

Determine the limiting reagent

Compare the mole ratio of the reacted substances to the balanced chemical equation: Ca₃(PO₄)₂: \(\frac{3.22 \text{ mol}}{2} = 1.61 \) H₂SO₄: \(\frac{9.99 \text{ mol}}{6} = 1.67 \) Since the Ca₃(PO₄)₂ mole ratio is smaller, Ca₃(PO₄)₂ is the limiting reagent.

Calculate the number of moles of the products formed

From the balanced chemical equation, we can see the mole ratio of the reactants with respect to the products: 1. Ca₃(PO₄)₂:CaSO₄ = 2:6 2. Ca₃(PO₄)₂:H₃PO₄ = 2:4 Given 3.22 mol Ca₃(PO₄)₂: - Moles of CaSO₄ produced = 6 × (3.22 mol Ca₃(PO₄)₂ / 2) = 9.66 mol CaSO₄ - Moles of H₃PO₄ produced = 4 × (3.22 mol Ca₃(PO₄)₂ / 2) = 6.44 mol H₃PO₄

Convert the moles of the products back into masses

We need the molar masses of the products: - CaSO₄: Molar mass = 40.08 + 32.07 + 4(16.00) ≈ 136.14 g/mol - H₃PO₄: Molar mass = 3(1.0079) + 31.00 + 4(16.00) ≈ 98.00 g/mol Now we can convert moles of CaSO₄ and H₃PO₄ back into their respective masses: - Mass of CaSO₄ produced = 9.66 mol × 136.14 g/mol ≈ 1315.87 g - Mass of H₃PO₄ produced = 6.44 mol × 98.00 g/mol ≈ 631.12 g Therefore, 1315.87 g of calcium sulfate (CaSO₄) and 631.12 g of phosphoric acid (H₃PO₄) can be produced from the reaction of 1.0 kg calcium phosphate with 1.0 kg concentrated sulfuric acid.

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantifiable relationships between the reactants and products in a chemical reaction. It is rooted in the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction. Consequently, the amount of each element must be the same in the products as it is in the reactants.

Understanding stoichiometry involves a series of steps. It begins with a balanced chemical equation, which ensures that the number of atoms for each element is the same on both sides of the reaction. Once the equation is balanced, you can use it as a map from which to determine the proportion of reactants and products involved.

In the provided exercise, stoichiometry helps us to utilize the balanced chemical equation to find out the exact amounts of calcium sulfate and phosphoric acid produced when 1.0 kg of calcium phosphate reacts with 1.0 kg of sulfuric acid.

Limiting Reagent

The concept of the limiting reagent is central to quantitative chemistry and stoichiometry. In essence, the limiting reagent is the substance that is used up first in a reaction, determining the maximum amount of product that can be formed. When it is exhausted, the reaction stops, regardless of how much of the other reactants are left.

Identifying the limiting reagent requires comparing the mole ratio of the available reactants to their ratios in the balanced chemical equation. In the textbook exercise, the chemical equation indicated that two moles of calcium phosphate could react with six moles of sulfuric acid to produce products. By calculating the amounts of each reagent available in moles, the equation showed that calcium phosphate was the limiting reagent because it had the smaller mole ratio. This analysis informs us that the amount of product will be based on the initial amount of calcium phosphate present.

Molar Mass Calculations

Molar mass calculations are critical for converting between the mass of a substance and the amount in moles, as moles are the units used in chemical equations. The molar mass is the weight of one mole of a substance and is measured in grams per mole (g/mol). It can be calculated by summing the atomic masses of all the atoms in a molecule as listed on the periodic table.

In the solution to the exercise, molar masses were necessary to convert the given masses (in kilograms) of calcium phosphate and sulfuric acid into moles. These mole quantities were then used to determine the mass of the products formed in the reaction. Without accurate molar mass calculations, it would not be possible to accurately predict the masses of products formed.