Final answer:
To prepare 5.0 tons of phosphorus, approximately 25.1 tons of Ca₃(PO₄)₂ are necessary.
Step-by-step explanation:
To determine the amount of Ca₃(PO₄)₂ necessary to prepare 5.0 tons of phosphorus, we first need to calculate the molar mass of Ca₃(PO₄)₂. The molar mass of Ca is 40.08 g/mol, the molar mass of P is 30.97 g/mol, and the molar mass of O is 16.00 g/mol. So, the molar mass of Ca₃(PO₄)₂ is (40.08 * 3) + (30.97 * 2) + (16.00 * 8) = 310.18 g/mol.
Next, we convert the given mass of phosphorus to moles. Since 1 ton = 1000 kg, 5.0 tons is equal to 5000 kg. Using the molar mass of phosphorus (30.97 g/mol), we can calculate the moles of phosphorus: (5000 kg / 1000) * (1 mole / 30.97 g) = 161.51 moles.
Now we can set up a stoichiometric ratio between Ca₃(PO₄)₂ and phosphorus. From the balanced equation, we know that 1 mole of Ca₃(PO₄)₂ produces 2 moles of phosphorus. Therefore, the required moles of Ca₃(PO₄)₂ is (161.51 moles / 2) = 80.75 moles.
Finally, to convert moles of Ca₃(PO₄)₂ to tons, we use the molar mass and conversion factor: 80.75 moles * (310.18 g / 1 mole) * (1 ton / 1000 kg) = 25.05 tons. Therefore, the answer is approximately 25.1 tons.