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Write a balanced equation for each of the following reactions: (a) Sulfur dioxide reacts with water. (b) Solid zinc sulfide reacts with hydrochloric acid. (c) Elemental sulfur reacts with sulfite ion to form thiosulfate. (d) Sulfur trioxide is dissolved in sulfuric acid.

Short Answer

Expert verified
(a) \( SO鈧 + H鈧侽 \rightarrow H鈧係O鈧 \) (b) \( ZnS + 2HCl \rightarrow ZnCl鈧 + H鈧係 \) (c) \( 8S鈧 + 8SO鈧兟测伝 \rightarrow 16S鈧侽鈧兟测伝 \) (d) \( SO鈧 + H鈧係O鈧 \rightarrow H鈧係鈧侽鈧 \)

Step by step solution

01

(a) Balancing the Reaction: Sulfur Dioxide with Water

Step 1: Identify reactants and products Sulfur dioxide (SO鈧) reacts with water (H鈧侽). Step 2: Write the unbalanced equation SO鈧 + H鈧侽 鈫 H鈧係O鈧 Step 3: Balance the equation The equation is already balanced: SO鈧 + H鈧侽 鈫 H鈧係O鈧
02

(b) Balancing the Reaction: Solid Zinc Sulfide with Hydrochloric Acid

Step 1: Identify reactants and products Solid zinc sulfide (ZnS) reacts with hydrochloric acid (HCl). Step 2: Write the unbalanced equation ZnS + HCl 鈫 ZnCl鈧 + H鈧係 Step 3: Balance the equation ZnS + 2HCl 鈫 ZnCl鈧 + H鈧係
03

(c) Balancing the Reaction: Elemental Sulfur with Sulfite Ion

Step 1: Identify reactants and products Elemental sulfur (S鈧) reacts with sulfite ions (SO鈧兟测伝) to form thiosulfate (S鈧侽鈧兟测伝). Step 2: Write the unbalanced equation S鈧 + SO鈧兟测伝 鈫 S鈧侽鈧兟测伝 Step 3: Balance the equation 8S鈧 + 8SO鈧兟测伝 鈫 16S鈧侽鈧兟测伝 Note: The coefficients are multiplied by 8 to account for the octasulfur molecule.
04

(d) Balancing the Reaction: Sulfur Trioxide with Sulfuric Acid

Step 1: Identify reactants and products Sulfur trioxide (SO鈧) is dissolved in sulfuric acid (H鈧係O鈧). Step 2: Write the unbalanced equation SO鈧 + H鈧係O鈧 鈫 H鈧係鈧侽鈧 Step 3: Balance the equation The equation is already balanced: SO鈧 + H鈧係O鈧 鈫 H鈧係鈧侽鈧

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Most popular questions from this chapter

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