Respuesta :

Space

Answer:

223 g KCl

General Formulas and Concepts:

Atomic Structure

  • Reading a Periodic Table
  • Moles

Stoichiometry

  • Using Dimensional Analysis
  • Analyzing reactions RxN

Explanation:

Step 1: Define

[RxN - Balanced] Cl₂ + 2KBr → Br₂ + 2KCl

[Given] 356 gg KBr

[Solve] g KCl

Step 2: Identify Conversions

[RxN] 2 mol KBr → 2 mol KCl

[PT] Molar Mass of K - 39.10 g/mol

[PT] Molar Mass of Br - 79.90 g/mol

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of KBr - 39.10 + 79.90 = 119 g/mol

Molar Mass of KCl - 39.10 + 35.45 = 74.55 g/mol

Step 3: Stoichiometry

  1. [DA] Set up conversion:                                                                                   [tex]\displaystyle 356 \ g \ KBr(\frac{1 \ mol \ KBr}{119 \ g \ KBr})(\frac{2 \ mol \ KCl}{2 \ mol \ KBr})(\frac{74.55 \ g \ KCl}{1 \ mol \ KCl})[/tex]
  2. [DA] Divide/Multiply [Cancel out units]:                                                           [tex]\displaystyle 223.024 \ g \ KCl[/tex]

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

223.024 g KCl ≈ 223 g KCl

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