Answer:
690 g AgCl
General Formulas and Concepts:
Math
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Chemistry
Atomic Structure
- Reading a Periodic Table
- Writing Compounds
Stoichiometry
- Using Dimensional Analysis
- Limiting Reactant/Excess Reactant
Step-by-step explanation:
Step 1: Define
[RxN - Unbalanced] AgNO₃ + ZnCl₂ → AgCl + Zn(NO₃)₂
↓
[RxN - Balanced] 2AgNO₃ + ZnCl₂ → 2AgCl + Zn(NO₃)₂
[Given] 2.4 mol ZnCl₂
[Solve] x g AgCl
Step 2: Identify Conversions
[RxN] 1 mol ZnCl₂ → 2 mol AgCl
[PT] Molar Mass of Ag - 107.87 g/mol
[PT] Molar Mass of Cl - 35.45 g/mol
Molar Mass of AgCl - 107.87 + 35.45 = 143.32 g/mol
Step 3: Stoich
- [DA] Set up:

- [DA] Multiply/Divide [Cancel out units]:

Step 4: Check
Follow sig fig rules and round. We are given 2 sig figs.
687.936 g AgCl ≈ 690 g AgCl