Problem 68 - Chemistry Central Science

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.

Step-by-Step Solution

Answer

(a) \(K_2O + H_2O \rightarrow 2KOH\) (b) \(P_4O_6 + 6H_2O \rightarrow 4H_3PO_3\) (c) \(Cr_2O_3 + 6HCl \rightarrow 2CrCl_3 + 3H_2O\) (d) \(SeO_2 + 2KOH \rightarrow K_2SeO_3 + H_2O\)

Step 1: 1. Reaction of potassium oxide with water

Potassium oxide (K2O) reacts with water (H2O) forming potassium hydroxide (KOH). Unbalanced equation: K2O + H2O -> KOH Now, we will balance this equation. Balanced equation: K2O + H2O -> 2KOH

Step 2: 2. Reaction of diphosphorus trioxide with water

Diphosphorus trioxide (P4O6) reacts with water (H2O) to form phosphoric acid (H3PO3). Unbalanced equation: P4O6 + H2O -> H3PO3 Now, we will balance this equation. Balanced equation: P4O6 + 6H2O -> 4H3PO3

Step 3: 3. Reaction of 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). Unbalanced equation: Cr2O3 + HCl -> CrCl3 + H2O Now, we will balance this equation. Balanced equation: Cr2O3 + 6HCl -> 2CrCl3 + 3H2O

Step 4: 4. Reaction of selenium dioxide with aqueous potassium hydroxide

Selenium dioxide (SeO2) reacts with aqueous potassium hydroxide (KOH) to form potassium selenite (K2SeO3) and water (H2O). Unbalanced equation: SeO2 + KOH -> K2SeO3 + H2O Now, we will balance this equation. Balanced equation: SeO2 + 2KOH -> K2SeO3 + H2O