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stoichiometry:
You conduct the following precipitation reaction in a lab:
CoCl₂ + 2NaOH → 2NaCl + Co(OH)₂
If you react 10.0 mL of 1.5 M CoCl₂ with plenty of NaOH, how many grams of Co(OH)₂ will precipitate out?

Respuesta :

Space

Answer:

1.4 g Co(OH)₂

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

Unit 0

  • Base 10 decimal system

Atomic Structure

  • Reading a Periodic Table
  • Molarity = moles of solute / liters of solution

Aqueous Solutions

  • States of Matter
  • Prediction Reactions RxN

Stoichiometry

  • Using Dimensional Analysis
  • Analyzing Reactions RxN

Explanation:

Step 1: Define

[RxN - Balanced] CoCl₂ + 2NaOH → 2NaCl + Co(OH)₂

[Given] 10.0 mL, 1.5 M CoCl₂

[Solve] grams Co(OH)₂

Step 2: Identify Conversions

[Base 10] 1000 mL = 1 L

[RxN] 1 mol CoCl₂ → 1 mol Co(OH)₂

[PT] Molar Mass of Co - 58.93 g/mol

[PT] Molar Mass of O - 16.00 g/mol

[PT] Molar Mass of H - 1.01 g/mol

Molar Mass of Co(OH)₂ - 58.93 + 2(16.00) + 2(1.01) = 92.95 g/mol

Step 3: Stoich

  1. [DA] Convert mL to L [Set up]:                                                                       [tex]\displaystyle 10.0 mL(\frac{1 \ L}{1000 \ mL})[/tex]
  2. [DA] Multiply/Divide [Cancel out units]:                                                           [tex]\displaystyle 0.01 \ L[/tex]
  3. [DA] Find moles of CoCl₂ [Molarity]:                                                              [tex]\displaystyle 1.5 \ M \ CoCl_2 = \frac{x \ mol \ CoCl_2}{0.01 \ L}[/tex]
  4. [DA] Solve for x [Multiplication Property of Equality]:                                   [tex]\displaystyle 0.015 \ mol \ CoCl_2[/tex]
  5. [DA] Set up [Reaction Stoich]:                                                                         [tex]\displaystyle 0.015 \ mol \ CoCl_2(\frac{1 \ mol \ Co(OH)_2}{1 \ mol \ CoCl_2})(\frac{92.95 \ g \ Co(OH)_2}{1 \ mol \ Co(OH)_2})[/tex]
  6. [DA] Multiply/Divide [Cancel out units]:                                                          [tex]\displaystyle 1.39425 \ g \ Co(OH)_2[/tex]

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs as our lowest.

1.39425 g Co(OH)₂ ≈ 1.4 g Co(OH)₂

Answer:

1.4 g Co(OH)₂

Explanation:

Molar mass of Co(OH)₂ = (58.9 + 16.0×2 + 1.0×2) g/mol = 92.9 g/mol

According to the given equation, mole ratio CoCl₂ : Co(OH)₂ = 1 : 1

No. of moles of CoCl₂ reacted = (1.5 mol/L) × (10.0/1000 mol) = 0.015 mol

No. of moles of Co(OH)₂ precipitated = 0.015 mol

Mass of Co(OH)₂ precipitated = (0.015 mol) × (92.9 g/mol) = 1.39 g / 1.4g

====

OR:

(1.5 mol CoCl₂ / 1000 mL CoCl₂ solution) × (10.0 mL CoCl₂ solution) × (1 mol Co(OH)₂ / 1 mol CoCl₂) × (92.9 g Co(OH)₂ / 1 mol Co(OH)₂)

= 1.39 g Co(OH)₂ / 1.4 g Co(OH)₂

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