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Chapter 7: Problem 66

Write balanced equations for the following reactions: (a) potassium oxide with water, (b) diphosphorus trioxide with water, (c) chromium(III) oxide with dilute hydrochloric acid, (d) selenium dioxide with aqueous potassium hydroxide.

Short Answer

Expert verified
a) \(K_2O + H_2O \rightarrow 2KOH\) b) \(P_2O_3 + 3H_2O \rightarrow 2H_3PO_3\) c) \(Cr_2O_3 + 6HCl \rightarrow 2CrCl_3 + 3H_2O\) d) \(SeO_2 + 2KOH \rightarrow K_2SeO_3 + H_2O\)

Step by step solution

01

a) Potassium oxide with water

First, we need to write down the reactants: potassium oxide (K₂O) and water (H₂O). Next, we identify the product: The potassium oxide will react with water to form potassium hydroxide (KOH). Now we balance the equation. We have 2 potassium ions on the left side, so we need 2 potassium ions on the right side, giving us: K₂O + H₂O → 2KOH
02

b) Diphosphorus trioxide with water

First, write down the reactants: diphosphorus trioxide (P₂O₃) and water (H₂O). Next, identify the product: Diphosphorus trioxide will react with water to form phosphorous acid (H₃PO₃). Now we balance the equation. We have 2 phosphorus and 3 oxygen atoms on both sides of the equation, so we need 6 hydrogen atoms on the right side: P₂O₃ + 3H₂O → 2H₃PO₃
03

c) Chromium(III) oxide with dilute hydrochloric acid

Write down the reactants: chromium(III) oxide (Cr₂O₃) and hydrochloric acid (HCl). Identify the products: Chromium(III) oxide reacts with hydrochloric acid to form chromium(III) chloride (CrCl₃) and water (H₂O). Next, we'll balance the equation. We have 2 chromium and 3 oxygen atoms on the left side, so we need 6 chlorine atoms on the right side to balance: Cr₂O₃ + 6HCl → 2CrCl₃ + 3H₂O
04

d) Selenium dioxide with aqueous potassium hydroxide

Write down the reactants: selenium dioxide (SeO₂) and potassium hydroxide (KOH). Identify the products: Selenium dioxide reacts with potassium hydroxide to form potassium selenite (K₂SeO₃) and water (H₂O). Now we balance the equation. We have 1 selenium and 2 oxygen atoms on both sides, so we need 2 potassium ions and 2 hydroxide ions on the right side: SeO₂ + 2KOH → K₂SeO₃ + H₂O

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Most popular questions from this chapter

The electron affinities, in \(\mathrm{kJ} / \mathrm{mol},\) for the group 1 \(\mathrm{B}\) and group 2 \(\mathrm{B}\) metals are as follows: (a) Why are the electron affinities of the group 2 \(\mathrm{B}\) elements greater than zero? (b) Why do the electron affinities of the group 1 \(\mathrm{B}\) elements become more negative as we move down the group? [Hint: Examine the trends in the electron affinities of other groups as we proceed down the periodic table.]

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