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A sample of iron (III) chloride has a mass of 26.29g. How many moles would this be?

User Putna
by
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1 Answer

9 votes

Answer:

0.1621 mol FeCl₃

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

Compounds

  • Determining compound formulas

Atomic Structure

  • Reading a Periodic Table
  • Using Dimensional Analysis

Step-by-step explanation:

Step 1: Define

26.29 g FeCl₃ (iron (III) chloride)

Step 2: Identify Conversions

Molar Mass of Fe - 55.85 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of FeCl₃ - 55.85 + 3(35.45) = 162.2 g/mol

Step 3: Convert

  1. Set up:
    \displaystyle 26.29 \ g \ FeCl_3((1 \ mol \ FeCl_3)/(162.2 \ g \ FeCl_3))
  2. Multiply:
    \displaystyle 0.162084 \ moles \ FeCl_3

Step 4: Check

Follow sig fig rules and round. We are given 4 sig figs.

0.162084 mol FeCl₃ ≈ 0.1621 mol FeCl₃

User El Padrino
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5.2k points