For the reaction between aqueous solutions of acetic acid (CH3COOH) and barium hydroxide, Ba(OH)2, 1. Write the balanced molecul
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<h3>Answer:</h3>
7.4797 g AlF₃
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
<u>Chemistry</u>
<u>Stoichiometry</u>
- Reading a Periodic Table
<u>Step 1: Define</u>
[RxN] 2AlF₃ + 3K₂O → 6KF + Al₂O₃
[Given] 15.524 g KF
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol AlF₃ = 6 mol KF
Molar Mass of Al - 26.98 g/mol
Molar Mass of F - 19.00 g/mol
Molar Mass of K - 39.10 g/mol
Molar Mass of AlF₃ - 26.98 + 3(19.00) = 83.98 g/mol
Molar Mass of KF - 39.10 + 19.00 = 58.10 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 5 sig figs.</em>
7.47966 g AlF₃ ≈ 7.4797 g AlF₃