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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: $$\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) + 3\mathrm{H}_{2}\mathrm{SO}_{4}(a q) \longrightarrow 3\mathrm{CaSO}_{4}(s) + 2\mathrm{H}_{3}\mathrm{PO}_{4}(a q)$$

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 \(0.98\mathrm{kg} \times \dfrac{1000\mathrm{g}}{1\mathrm{kg}} \times \dfrac{1 \: mol}{98.08 \: g}= \dfrac{980}{98.08} \: mol = 9.99 \: mol\)

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.

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

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.