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Chapter 5: Problem 112

Phosphorus can be prepared from calcium phosphate by the following reaction: $$\begin{aligned}2 \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+6 \mathrm{SiO}_{2}(s)+& 10 \mathrm{C}(s) \longrightarrow \\\& 6 \mathrm{CaSiO}_{3}(s)+\mathrm{P}_{4}(s)+10 \mathrm{CO}(g) \end{aligned}$$ 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 P₄ that can be produced from 1.0 kg of phosphorite is approximately \(149.79 \, \text{g}\).

Step by step solution

01

Calculate the mass of calcium phosphate in the phosphorite sample

Since the phosphorite sample is 75% calcium phosphate by mass, we can calculate the mass of calcium phosphate in the 1.0 kg phosphorite sample as follows: Mass of calcium phosphate = (1.0 kg phosphorite) × (0.75) = 0.75 kg Since we will be dealing with grams throughout the problem, let's convert the mass to grams: 0.75 kg = 750 g
02

Determine the moles of calcium phosphate

To determine the moles of calcium phosphate, we need to find the molar mass of Ca₃(PO₄)₂: Molar mass of Ca₃(PO₄)₂ = (3 × molar mass of Ca) + (2 × molar mass of P) + (8 × molar mass of O) = (3 × 40.08 g/mol) + (2 × 30.97 g/mol) + (8 × 16.00 g/mol) = 310.18 g/mol Now, let's calculate the moles of calcium phosphate in the sample: Moles of Ca₃(PO₄)₂ = (750 g Ca₃(PO₄)₂) / (310.18 g/mol) = 2.418 mol
03

Use stoichiometry to calculate the moles of P₄ produced

The balanced chemical equation gives us the stoichiometric ratio between the moles of Ca₃(PO₄)₂ and moles of P₄ produced: 2 moles of Ca₃(PO₄)₂ → 1 mole of P₄ Using this ratio, we can calculate the moles of P₄ produced from the moles of Ca₃(PO₄)₂: Moles of P₄ = (1 mol P₄ / 2 mol Ca₃(PO₄)₂) × (2.418 mol Ca₃(PO₄)₂) = 1.209 mol P₄
04

Convert the moles of P₄ to grams

To find the mass of P₄ produced, we need to find the molar mass of P₄: Molar mass of P₄ = 4 × molar mass of P = 4 × 30.97 g/mol = 123.88 g/mol Now, let's calculate the grams of P₄ produced: Mass of P₄ = (1.209 mol P₄) × (123.88 g/mol) = 149.79 g The maximum amount of P₄ that can be produced from 1.0 kg of phosphorite is approximately 149.79 g.

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Most popular questions from this chapter

An element X forms both a dichloride (XCI_) and a tetrachloride \(\left(\mathrm{XCl}_{4}\right) .\) Treatment of \(10.00 \mathrm{g} \mathrm{XCl}_{2}\) with excess chlorine forms \(12.55 \mathrm{g} \mathrm{XCl}_{4} .\) Calculate the atomic mass of \(\mathrm{X},\) and identify \(\mathrm{X}\).

Bornite \(\left(\mathrm{Cu}_{3} \mathrm{FeS}_{3}\right)\) is a copper ore used in the production of copper. When heated, the following reaction occurs: $$2 \mathrm{Cu}_{3} \mathrm{FeS}_{3}(s)+7 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{Cu}(s)+2 \mathrm{FeO}(s)+6 \mathrm{SO}_{2}(g)$$ If 2.50 metric tons of bornite is reacted with excess \(\mathrm{O}_{2}\) and the process has an \(86.3 \%\) yield of copper, what mass of copper is produced?

One of the components that make up common table sugar is fructose, a compound that contains only carbon, hydrogen, and oxygen. Complete combustion of \(1.50 \mathrm{g}\) of fructose produced \(2.20 \mathrm{g}\) of carbon dioxide and \(0.900 \mathrm{g}\) of water. What is the empirical formula of fructose?

Consider the following unbalanced reaction: $$\mathrm{P}_{4}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{PF}_{3}(g)$$ What mass of \(\mathrm{F}_{2}\) is needed to produce \(120 . \mathrm{g}\) of \(\mathrm{PF}_{3}\) if the reaction has a \(78.1 \%\) yield?

Consider the following unbalanced equation: $$\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{CaSO}_{4}(s)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q)$$ What masses of calcium sulfate and phosphoric acid can be produced from the reaction of \(1.0 \mathrm{kg}\) calcium phosphate with \(1.0 \mathrm{kg}\) concentrated sulfuric acid \(\left(98 \% \mathrm{H}_{2} \mathrm{SO}_{4}\text { by mass)? }\right.\)
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