Write a balanced equation for each of the following reactions: (a) hydrolysis of \(\mathrm{PCl}_{5},(\mathbf{b})\) dehydration of phosphoric acid (also called orthophosphoric acid) to form pyrophosphoric acid, \((\mathbf{c})\) reaction of \(\mathrm{P}_{4} \mathrm{O}_{10}\) with water.

Short Answer

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(a) Hydrolysis of PCl5: \[ PCl_{5 (s)} + 4H_{2 (l)}O \rightarrow H_{3 (aq)}PO_{4} + 5H_{(aq)}Cl \] (b) Dehydration of phosphoric acid to form pyrophosphoric acid: \[ 2H_{3 (aq)}PO_{4} \rightarrow H_{4 (aq)}P_{2}O_{7} + H_{2 (l)}O \] (c) Reaction of P4O10 with water: \[ P_{4 (s)}O_{10} + 6H_{2 (l)}O \rightarrow 4H_{3 (aq)}PO_{4} \]

Step by step solution

01

(a) Hydrolysis of PCl5

Hydrolysis of a compound involves the reaction of a substance with water, which results in the splitting of the compound into ions or simpler molecules. In the case of PCl5, it reacts with water (H2O) to form phosphoric acid (H3PO4) and hydrochloric acid (HCl) as products. The balanced equation for this reaction is: \[ PCl_{5 (s)} + 4H_{2 (l)}O \rightarrow H_{3 (aq)}PO_{4} + 5H_{(aq)}Cl \]
02

(b) Dehydration of phosphoric acid to form pyrophosphoric acid

Dehydration is the process of removing water from a compound. In this case, we need to remove water from two molecules of phosphoric acid (H3PO4) to form one molecule of pyrophosphoric acid (H4P2O7). The balanced equation for this reaction is: \[ 2H_{3 (aq)}PO_{4} \rightarrow H_{4 (aq)}P_{2}O_{7} + H_{2 (l)}O \]
03

(c) Reaction of P4O10 with water

The reaction of P4O10 with water involves the formation of phosphoric acid (H3PO4) as the product from tetraphosphorus decoxide. The balanced equation for this reaction is: \[ P_{4 (s)}O_{10} + 6H_{2 (l)}O \rightarrow 4H_{3 (aq)}PO_{4} \] These balanced equations show the stoichiometry of each reaction, ensuring that the number of atoms of each element on the reacting side equals the number of atoms of the same element on the product side.

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Most popular questions from this chapter

The physical properties of \(\mathrm{D}_{2} \mathrm{O}\) differ from those of \(\mathrm{H}_{2} \mathrm{O}\) because (a) D has a different electron configuration than O. (b) \(\mathrm{D}\) is radioactive. (c) D forms stronger bonds with \(\mathrm{O}\) than \(\mathrm{H}\) does. (d) D is much more massive than H.

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