91Ó°ÊÓ

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 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 = \(\frac{0.98 \times 1000\, \text{g}}{98.08\, \text{g/mol}} = 9.986\, \text{mol}\)

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.

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.