How much of a 0.250 M sucrose solution must be used to prepare 400.0 mL of a 0.0310 M solution
2 answers:
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<h3>Answer:</h3>
4 g AgCl
<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
- Using Dimensional Analysis
<u>Step 1: Define</u>
[RxN] 2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂
[Given] 5.0 g AgNO₃
<u>Step 2: Identify Conversions</u>
[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
<u>Step 3: Stoichiometry</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 1 sig fig.</em>
4.21533 g AgCl ≈ 4 g AgCl