Respuesta :

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Answer:

7.4797 g AlF₃

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] 2AlF₃ + 3K₂O → 6KF + Al₂O₃

[Given] 15.524 g KF

Step 2: Identify Conversions

[RxN] 6 mol KF = 2 mol AlF₃

Molar Mass of K - 39.10 g/mol

Molar Mass of F - 19.00 g/mol

Molar Mass of Al - 26.98 g/mol

Molar Mass of KF - 39.10 + 19.00 = 58.1 g/mol

Molar Mass of AlF₃ - 26.98 + 3(19.00) = 83.98 g/mol

Step 3: Stoichiometry

  1. Set up:                              [tex]\displaystyle 15.524 \ g \ KF(\frac{1 \ mol \ KF}{58.1 \ g \ KF})(\frac{2 \ mol \ AlF_3}{6 \ mol \ KF})(\frac{83.98 \ g \ AlF_3}{1 \ mol \ AlF_3})[/tex]
  2. Multiply/Divide:                                                                                                  [tex]\displaystyle 7.47966 \ g \ AlF_3[/tex]

Step 4: Check

Follow sig fig rules and round. We are given 5 sig figs.

7.47966 g AlF₃ ≈ 7.4797 g AlF₃

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