Phosphoric acid can be prepared by reaction of sulfuric acid with "phosphate rock" according to the equation: Ca₃(PO₄) ₂ + 3H₂SO₄ $\to$ 3CaSO₄ + 2H₃PO₄
Suppose the reaction is carried out starting with 129 g of Ca₃(PO₄) ₂ and 97.4 g of H₂SO₄.Which substance is the limiting reactant?

Question

Phosphoric acid can be prepared by reaction of sulfuric acid with "phosphate rock" according to the equation: Ca₃(PO₄) ₂ + 3H₂SO₄

\to

3CaSO₄ + 2H₃PO₄
Suppose the reaction is carried out starting with 129 g of Ca₃(PO₄) ₂ and 97.4 g of H₂SO₄.Which substance is the limiting reactant?

Quizplus · Accepted Answer

To determine the limiting reactant, we need to calculate the moles of each reactant. Firstly, we need to convert the given masses to moles using their respective molar masses. The molar mass of Ca₃(PO₄)₂ is calculated as follows: 1 atomic mass unit (u) = 1 g/mol 1 Ca = 40.078 u = 40.078 g/mol 1 P = 30.974 u = 30.974 g/mol 1 O (x14) = 15.999 u (x14) = 223.986 g/mol 1 S (x1) = 32.06 u (x1) = 32.06 g/mol Total molar mass of Ca₃(PO₄)₂ = 327.098 g/mol Moles of Ca₃(PO₄)₂ = Mass ÷ Molar mass = 129 g ÷ 327.098 g/mol ≈ 0.394 mol Similarly, the molar mass of H₂SO₄ is calculated as follows: 1 H = 1.008 u = 1.008 g/mol 1 S = 32.06 u = 32.06 g/mol 1 O (x4) = 15.999 u (x4) = 63.996 g/mol 1 P = 30.974 u = 30.974 g/mol Total molar mass of H₂SO₄ = 128.038 g/mol Moles of H₂SO₄ = Mass ÷ Molar mass = 97.4 g ÷ 128.038 g/mol ≈ 0.761 mol We can see from the balanced chemical equation that the stoichiometric ratio of Ca₃(PO₄)₂ to H₂SO₄ is 1:3. That means for every 1 mole of Ca₃(PO₄)₂, we need 3 moles of H₂SO₄. Moles required of H₂SO₄ = (3 moles of Ca₃(PO₄)₂) x (0.394 mol) = 1.182 mol Since we only have 0.761 mol of H₂SO₄, it is the limiting reactant. Therefore, the answer is B.

Phosphoric Acid Can Be Prepared by Reaction of Sulfuric Acid →\to→

Phosphoric Acid Can Be Prepared by Reaction of Sulfuric Acid $\to$