Question

For the reaction represented by the equation:

Cl2 + 2KBr → Br2 + 2KCl

how many grams of KCl can be produced from 356 grams KBr?

Answer 1

223 g KCl

General Formulas and Concepts:

Atomic Structure

• Reading a Periodic Table
• Moles

Stoichiometry

• Using Dimensional Analysis
• Analyzing reactions RxN

Step-by-step 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:
   `\displaystyle 356 \ g \ KBr((1 \ mol \ KBr)/(119 \ g \ KBr))((2 \ mol \ KCl)/(2 \ mol \ KBr))((74.55 \ g \ KCl)/(1 \ mol \ KCl))`
2. [DA] Divide/Multiply [Cancel out units]:
   `\displaystyle 223.024 \ g \ KCl`

Step 4: Check

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

223.024 g KCl ≈ 223 g KCl