Question

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

The balanced equations for the given reactions are: a) $K_{2}O+H_{2}O\to 2KOH$ b) $P_4{O}_6+6H_{2}O\to 4H_{3}P{O}_3$ c) $C{r}_2{O}_3+6HCl\to 2CrC{l}_3+3H_{2}O$ d) $Se{O}_2+2KOH\to K_{2}Se{O}_3+H_{2}O$

Step-by-step solution

a) Potassium oxide with water

Potassium oxide (K2O) reacts with water (H2O) to produce potassium hydroxide (KOH). The balanced equation for this reaction can be found by ensuring that the number of atoms of each element is equal on both sides of the equation: $$K_{2}O+H_{2}O\to 2KOH$$

b) Diphosphorus trioxide with water

Diphosphorus trioxide (P4O6) reacts with water (H2O) to form phosphorous acid (H3PO3). To balance the equation, ensure the number of atoms of each element is equal on both sides: $$P_4{O}_6+6H_{2}O\to 4H_{3}P{O}_3$$

c) Chromium(III) oxide with dilute hydrochloric acid

Chromium(III) oxide (Cr2O3) reacts with dilute hydrochloric acid (HCl) to form chromium(III) chloride (CrCl3) and water (H2O). Balance the equation to ensure the same number of atoms of each element on both sides: $$C{r}_2{O}_3+6HCl\to 2CrC{l}_3+3H_{2}O$$

d) Selenium dioxide with aqueous potassium hydroxide

Selenium dioxide (SeO2) reacts with aqueous potassium hydroxide (KOH) to form potassium selenite (K2SeO3) and water (H2O). Ensure the number of atoms of each element is equal on both sides to balance the equation: $$Se{O}_2+2KOH\to K_{2}Se{O}_3+H_{2}O$$

Key concepts

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry. This process ensures that the number of atoms for each element is the same on both sides of a chemical equation. A balanced equation respects the law of conservation of mass, which states that matter cannot be created or destroyed. When balancing equations:

• Identify each type of atom in the reactants and products.
• Count the atoms of each element on both sides of the equation.
• Add coefficients (whole numbers) in front of formulas to equalize the number of atoms for each element on both sides.

For instance, in the reaction between potassium oxide and water: $$K_{2}O+H_{2}O\to 2KOH$$Ensure the atoms equal:

• K (Potassium): 2 on both sides
• O (Oxygen): 2 on both sides
• H (Hydrogen): 2 on both sides

The balanced equation shows 1 molecule of water reacts with 1 molecule of potassium oxide to produce 2 molecules of potassium hydroxide.

Inorganic Chemistry

Inorganic chemistry focuses on compounds and elements, excluding organic compounds. It often involves studying reactions between metals, nonmetals, acids, and bases.In the example of chromium(III) oxide reacting with hydrochloric acid:$$C{r}_2{O}_3+6HCl\to 2CrC{l}_3+3H_{2}O$$Here we see:

• Chromium(III) oxide, a metal oxide, acts as a source of Cr³⁺ ions.
• Hydrochloric acid provides H⁺ and Cl^- ions.

Together, they form chromium(III) chloride and water. Balancing these reactions involves ensuring that the oxidation states of the elements are correctly accounted for, and both mass and charge balance are achieved.

Acid-Base Reactions

Acid-base reactions are central in chemistry, where an acid reacts with a base to produce a salt and water. These reactions are characterized by the transfer of protons (H⁺ ions).Consider the reaction of selenium dioxide with potassium hydroxide:$$Se{O}_2+2KOH\to K_{2}Se{O}_3+H_{2}O$$Here's what happens:

• Potassium hydroxide ($KOH$) is a strong base, providing OH^- ions.
• Selenium dioxide ($Se{O}_2$) reacts with the base to form potassium selenite ($K_{2}Se{O}_3$), a salt, and water.

This is a typical neutralization reaction where the acid in the oxide form reacts with the base to form a salt and water, showcasing the fundamental principles of acid-base chemistry.