What volume in milliliters of 0.0150 M Ba(OH)2 is required to neutralize 50.0 mL of 0.0500 M HBr?
1 answer:
The volume of required will be 83.33 mL
The equation of the reaction is as follows:
Mole ratio of reactants = 1:2
Mole of 50 mL, 0.05 M HBr = 0.05 x 0.05 = 0.0025 moles
Equivalent mole of = 0.0025/2 = 0.00125 moles
Volume of 0.015 M, 0.00125 moles = 0.00125/0.015 = 0.0833 L
0.0833 L = 83.33 mL
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<h3>Answer:</h3>
128 g HCl
<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> </u>
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
<u>Stoichiometry</u>
- Reaction Mole Ratios
- Using Dimensional Analysis
<u>Step 1: Define</u>
[RxN - Unbalanced] Mg (s) + HCl (aq) → MgCl (aq) + H₂ (g)
↓
[RxN - Balanced] 2Mg (s) + 2HCl (aq) → 2MgCl (aq) + H₂ (g)
[Given] 3.25 mol Mg
[Solve] x g HCl
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol Mg → 2 mol HCl
[PT] Molar Mass of H - 1.01 g/mol
[PT] Molar Mass of Cl - 35.45 g/mol
Molar Mass of HCl - 1.01 + 35.45 = 36.46 g/mol
<u>Step 3: Stoich</u>
- [S - DA] Set up:
- [S - DA] Multiply/Divide [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
127.61 g HCl ≈ 128 g HCl