Question

2AlF3 + 3K2O → 6KF + Al2O3

How many grams of AlF3would it take to make 15.524 g of KF?

?g AlF3 (use 5 sig figs)

Answer 1

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

Step-by-step 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:
   `\displaystyle 15.524 \ g \ KF((1 \ mol \ KF)/(58.1 \ g \ KF))((2 \ mol \ AlF_3)/(6 \ mol \ KF))((83.98 \ g \ AlF_3)/(1 \ mol \ AlF_3))`
2. Multiply/Divide:
   `\displaystyle 7.47966 \ g \ AlF_3`

Step 4: Check

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

7.47966 g AlF₃ ≈ 7.4797 g AlF₃