Respuesta :

Space

Answer:

4 g AgCl

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. 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

  1. 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]
  2. 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

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