Question

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?

Answer 1

1.4 g Co(OH)₂

Step-by-step 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

Answer 2

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

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

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