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Chapter 3: Problem 118

Phosphorus can be prepared from calcium phosphate by the following reaction: $$ 2 \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+6 \mathrm{SiO}_{2}(s)+10 \mathrm{C}(s) \longrightarrow $$ $$ 6 \mathrm{CaSiO}_{3}(s)+\mathrm{P}_{4}(s)+10 \mathrm{CO}(g) $$ Phosphorite is a mineral that contains \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) plus other non-phosphorus-containing compounds. What is the maximum amount of \(\mathrm{P}_{4}\) that can be produced from 1.0 \(\mathrm{kg}\) of phosphorite if the phorphorite sample is 75$\% \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}$ by mass? Assume an excess of the other reactants.

Short Answer

Expert verified
The maximum amount of phosphorus (P₄) that can be produced from 1.0 kg of phosphorite containing 75% calcium phosphate by mass is approximately 49.88 g.

Step by step solution

01

Calculate the mass of calcium phosphate

Since the given phosphorite sample contains 75% calcium phosphate by mass, the mass of calcium phosphate in the 1.0 kg sample can be calculated as follows: Mass of calcium phosphate = 75% of 1.0 kg = 0.75 kg.
02

Determine the number of moles of calcium phosphate

To determine the number of moles of calcium phosphate, divide the mass of calcium phosphate by its molar mass. The molar mass of Ca₃(PO₄)₂ is given by: Molar mass of Ca₃(PO₄)₂ = 3 × (mass of Ca) + 2 × (mass of P + 4 × mass of O). Plugging in the values from the periodic table: - Mass of Ca = 40.08 g/mol - Mass of P = 30.97 g/mol - Mass of O = 16.00 g/mol Molar mass of Ca₃(PO₄)₂ = 3 × 40.08 + 2 × (30.97 + 4 × 16.00) = 310.18(3) + 30.97(2) + 16.00(8) = 930.54 g/mol Now, calculate the number of moles of calcium phosphate: Moles of Ca₃(PO₄)₂ = \(\frac{0.75\ kg}{930.54\ \frac{g}{mol}}\) = \(\frac{750\ g}{930.54\ \frac{g}{mol}}\) = 0.8061 mol (4 s.f.)
03

Use stoichiometry to determine the number of moles of phosphorus (P₄) that can be produced

The balanced chemical equation is: 2 Ca₃(PO₄)₂ + 6 SiO₂ + 10 C → 6 CaSiO₃ + P₄ + 10 CO From this equation, we can see that 2 moles of calcium phosphate yield 1 mole of phosphorus (P₄). Therefore, the number of moles of phosphorus (P₄) that can be produced is given by: Moles of P₄ = \(\frac{1}{2}\) × moles of Ca₃(PO₄)₂ = \(\frac{1}{2}\) × 0.8061 mol = 0.40305 mol (5 s.f.)
04

Calculate the mass of phosphorus that can be produced

To find the mass of phosphorus that can be produced, multiply the number of moles of P₄ by its molar mass. The molar mass of P₄ is given by: Molar mass of P₄ = 4 × mass of P = 4 × 30.97 g/mol = 123.88 g/mol Now, calculate the mass of phosphorus that can be produced: Mass of P₄ = moles of P₄ × molar mass of P₄ = 0.40305 mol × 123.88 g/mol = 49.88 g (4 s.f.) Therefore, the maximum amount of phosphorus (P₄) that can be produced from 1.0 kg of phosphorite containing 75% calcium phosphate by mass is approximately 49.88 g.

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