Write a balanced equation for each of the following reactions: (a) preparation of white phosphorus from calcium phosphate, (b) hydrolysis of \(\mathrm{PBr}_{3}\), (c) reduction of \(\mathrm{PBr}_{3}\) to \(\mathrm{P}_{4}\) in the gas phase, using \(\mathrm{H}_{2}\).

Short Answer

Expert verified
The short answers are as follows: (a) 10 Ca3(PO4)2 + 30 C + 6 SiO2 -> 2 P4 + 30 CaSiO3 + 30 CO (b) \( \mathrm{PBr}_{3} \) + 3 H2O -> H3PO3 + 3 HBr (c) 4 \( \mathrm{PBr}_{3} \) + 6 H2 -> P4 + 12 HBr

Step by step solution

01

Reaction (a): Preparation of white phosphorus from calcium phosphate

For this reaction, we start with calcium phosphate (Ca3(PO4)2) and use elemental carbon (C) and silicon dioxide (SiO2) as reducing agents. The reaction produces white phosphorus (P4), calcium silicate (CaSiO3), and carbon monoxide (CO). To balance this equation, we need to ensure the number of atoms for each element on both sides of the equation is the same. Ca3(PO4)2 + C + SiO2 -> P4 + CaSiO3 + CO There are 6 oxygen atoms on the reactant side, so we need 6 oxygen atoms on the product side. Add coefficients to balance the equation: 10 Ca3(PO4)2 + 30 C + 6 SiO2 -> 2 P4 + 30 CaSiO3 + 30 CO This balanced equation reflects the stoichiometry of the reactants and products in the reaction.
02

Reaction (b): Hydrolysis of \(\mathrm{PBr}_{3}\)

For this reaction, we have phosphorus tribromide (\(\mathrm{PBr}_{3}\)) reacting with water (H2O) to form phosphorous acid (H3PO3) and hydrogen bromide (HBr). To balance this equation, we need to ensure the number of atoms for each element on both sides of the equation is the same. \(\mathrm{PBr}_{3}\) + H2O -> H3PO3 + HBr There are 3 bromine atoms on the reactant side, so we need 3 bromine atoms on the product side. And we need to balance the number of hydrogen atoms on both sides: \(\mathrm{PBr}_{3}\) + 3 H2O -> H3PO3 + 3 HBr The balanced equation represents the correct stoichiometry for this hydrolysis reaction.
03

Reaction (c): Reduction of \(\mathrm{PBr}_{3}\) to \(\mathrm{P}_{4}\) in the gas phase using \(\mathrm{H}_{2}\)

For this reaction, phosphorus tribromide (\(\mathrm{PBr}_{3}\)) in the gas phase reacts with hydrogen gas (H2) to form white phosphorus (P4) and hydrogen bromide (HBr) as products. To balance this equation, we need to ensure the number of atoms for each element on both sides of the equation is the same. \(\mathrm{PBr}_{3}\) + H2 -> P4 + HBr There are 4 phosphorus atoms on the product side, so we need 4 phosphorus atoms on the reactant side. Similarly, we need to balance the number of bromine and hydrogen atoms: 4 \(\mathrm{PBr}_{3}\) + 6 H2 -> P4 + 12 HBr Now the balanced equation represents the stoichiometry of the reactants and products in this reduction reaction.

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Most popular questions from this chapter

The standard heats of formation of $\mathrm{H}_{2} \mathrm{O}(g), \mathrm{H}_{2} \mathrm{~S}(g), \mathrm{H}_{2} \operatorname{Se}(g)$, and \(\mathrm{H}_{2} \mathrm{Te}(g)\) are \(-241.8,-20.17,+29.7,\) and $+99.6 \mathrm{~kJ} /$ mol, respectively. The enthalpies necessary to convert the elements in their standard states to one mole of gaseous atoms are \(248,277,227,\) and \(197 \mathrm{~kJ} / \mathrm{mol}\) of atoms for $\mathrm{O}, \mathrm{S}$, Se, and Te, respectively. The enthalpy for dissociation of \(\mathrm{H}_{2}\) is \(436 \mathrm{~kJ} / \mathrm{mol}\). Calculate the average \(\mathrm{H}-\mathrm{O}, \mathrm{H}-\mathrm{S}, \mathrm{H}-\mathrm{Se}\), and \(\mathrm{H}-\) Te bond enthalpies, and comment on their trend.

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