Answer :
Answer:
4 g AgCl
General Formulas and Concepts:
Math
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
Chemistry
Stoichiometry
- Reading a Periodic Table
- Using Dimensional Analysis
Explanation:
Step 1: Define
[RxN] 2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂
[Given] 5.0 g AgNO₃
Step 2: Identify Conversions
[Reaction - Stoich] 2AgNO₃ → 2AgCl
Molar Mass of Ag - 107.87 g/mol
Molar Mass of N - 14.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Cl - 35.45 g/mol
Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol
Molar Mass of AgCl - 107.87 + 35.45 = 143.32 g/mol
Step 3: Stoichiometry
- Set up: [tex]\displaystyle 5.0 \ g \ AgNO_3(\frac{1 \ mol \ AgNO_3}{169.88 \ g \ AgNO_3})(\frac{2 \ mol \ AgCl}{2 \ mol \ AgNO_3})(\frac{143.22 \ g \ AgCl}{1 \ mol \ AgCl})[/tex]
- Multiply/Divide: [tex]\displaystyle 4.21533 \ g \ AgCl[/tex]
Step 4: Check
Follow sig fig rules and round. We are given 1 sig fig.
4.21533 g AgCl ≈ 4 g AgCl