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Chapter 3: Problem 104

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 nonphosphorus- 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, which is 75% Ca3(PO4)2 by mass, is 149.78g.

Step by step solution

01

Calculate the moles of the limiting reactant Ca3(PO4)2 in the sample.

In this case, the limiting reactant is Ca3(PO4)2, as it is the only reactant affected by the mass of the phosphorite. We are given that the mineral is 75% Ca3(PO4)2 by mass, so we first find the mass of Ca3(PO4)2 in the 1 kg sample: Mass of Ca3(PO4)2 = (75/100) * 1000g = 750g Next, determine the molar mass of Ca3(PO4)2: Molar mass of Ca3(PO4)2 = (3 * 40.08g/mol Ca) + (2 * (1 * (30.97g/mol P) + 4 * (16g/mol O))) = 310.18g/mol Now, we can convert the mass of Ca3(PO4)2 to moles: Moles of Ca3(PO4)2 = 750g / 310.18g/mol = 2.418 mol
02

Use stoichiometry to determine the moles of P4 produced.

Based on the balanced chemical equation, we know that 2 moles of Ca3(PO4)2 produce 1 mole of P4. So we can set up a proportion to calculate the number of moles of P4 produced from the moles of Ca3(PO4)2: (1 mol P4 / 2 mol Ca3(PO4)2) = (x moles P4 / 2.418 mol Ca3(PO4)2) Cross-multiplying and solving for x, we get: x moles P4 = (1 mol P4) * (2.418 mol Ca3(PO4)2) / 2 mol Ca3(PO4)2 x moles P4 = 1.209 mol
03

Calculate the mass of P4 produced from the moles of P4.

Since we now have the moles of P4 produced, we can calculate the mass of P4 using its molar mass: Molar mass of P4 = 4 * (30.97g/mol P) = 123.88g/mol So, mass of P4 produced = 1.209 mol * 123.88g/mol = 149.78g Hence, the maximum amount of P4 that can be produced from 1.0 kg of phosphorite is 149.78g.

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Most popular questions from this chapter

A \(2.077-g\) sample of an element, which has an atomic mass between 40 and 55 , reacts with oxygen to form \(3.708 \mathrm{~g}\) of an oxide. Determine the formula of the oxide (and identify the element).

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Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane: $$ 2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g)+2 \mathrm{CH}_{4}(g) \longrightarrow 2 \mathrm{HCN}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ If \(5.00 \times 10^{3} \mathrm{~kg}\) each of \(\mathrm{NH}_{3}, \mathrm{O}_{2}\), and \(\mathrm{CH}_{4}\) are reacted, what mass of \(\mathrm{HCN}\) and of \(\mathrm{H}_{2} \mathrm{O}\) will be produced, assuming \(100 \%\) yield?

Ammonia reacts with \(\mathrm{O}_{2}\) to form either \(\mathrm{NO}(\mathrm{g})\) or \(\mathrm{NO}_{2}(\mathrm{~g})\) according to these unbalanced equations: $$ \begin{array}{l} \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{array} $$ In a certain experiment \(2.00 \mathrm{~mol} \mathrm{NH}_{3}(g)\) and \(10.00 \mathrm{~mol}\) \(\mathrm{O}_{2}(g)\) are contained in a closed flask. After the reaction is complete, \(6.75 \mathrm{~mol} \mathrm{O}_{2}(g)\) remains. Calculate the number of moles of \(\mathrm{NO}(g)\) in the product mixture: (Hint: You cannot do this problem by adding the balanced equations, because you cannot assume that the two reactions will occur with equal probability.)

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