Respuesta :

Space

Answer:

200 g CuSO₄ · 5H₂O

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

Atomic Structure

  • Reading a Periodic Table
  • Moles
  • Compounds

Organic

  • Prefix roots

Reactions

  • Salts
  • Anhydride Salts
  • Hydrates

Stoichiometry

  • Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 1 mol CuSO₄ · 5H₂O (Copper II sulfate pentahydrate)

Step 2: Identify Conversions

[PT] Molar Mass of Cu - 63.55 g/mol

[PT] Molar Mass of S - 32.07 g/mol

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

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

Molar Mass of CuSO₄ - 63.55 + 32.07 + 4(16.00) = 159.62 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Step 3: Convert

There is 5 moles of water added for every one mole of Copper II Sulfate present.

  1. [DA] Set up:                                                                                                     [tex]\displaystyle 1 \ mol \ CuSO_4(\frac{159.62 \ g \ CuSO_4}{1 \ mol \ CuSO_4}) + 5 \ mol \ H_2O(\frac{18.02 \ g \ H_2O}{1 \ mol \ H_2O})[/tex]
  2. [DA] Multiply/Divide [Cancel out units]:                                                         [tex]\displaystyle 159.62 \ g \ CuSO_4 + 90.1 \ g \ H_2O[/tex]
  3. [DA] Add:                                                                                                         [tex]\displaystyle 249.72 \ g \ CuSO_4 \cdot 5H_2O[/tex]

Step 4: Check

Follow sig fig rules and round. We are given 1 sig fig.

249.72 g CuSO₄ · 5H₂O ≈ 200 g CuSO₄ · 5H₂O