Question

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

Answer 1

Answer:

0.1621 mol FeCl₃

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

Set up: \displaystyle 26.29 \ g \ FeCl_3(\frac{1 \ mol \ FeCl_3}{162.2 \ g \ FeCl_3})26.29 g FeCl

3

(

162.2 g FeCl

3

1 mol FeCl

3

)

Multiply: \displaystyle 0.162084 \ moles \ FeCl_30.162084 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