Answer:
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
- [DA] Set up conversion:
- [DA] Divide/Multiply [Cancel out units]:
Step 4: Check
Follow sig fig rules and round. We are given 3 sig figs.
223.024 g KCl ≈ 223 g KCl