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2 years ago

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

Chemistry

1 answer:

vekshin12 years ago

8 0

<h3>Answer:</h3>

0.1621 mol FeCl₃

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Compounds</u>

Determining compound formulas

<u>Atomic Structure</u>

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

26.29 g FeCl₃ (iron (III) chloride)

<u>Step 2: Identify Conversions</u>

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

<u>Step 3: Convert</u>

Set up: $\displaystyle 26.29 \ g \ FeCl_3(\frac{1 \ mol \ FeCl_3}{162.2 \ g \ FeCl_3})$
Multiply: $\displaystyle 0.162084 \ moles \ FeCl_3$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 4 sig figs.</em>

0.162084 mol FeCl₃ ≈ 0.1621 mol FeCl₃

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Answer:

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ΔGrxn = ∑Gproducts - ∑Greactants

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M₂O₃ = -6.7 kJ/mol

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$lnK=-\frac{deltaG_{rxn} }{RT}$

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ΔGrxn = 6.7 kJ/mol = 6700 J/mol

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$lnK=-\frac{6700}{8.314*298} =-2.704\\K=0.066$

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Explanation:

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

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