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HELP!!!!
Iron (III) oxide reacts with solid carbon in the followed reaction: 2Fe2O3(s) + 3C(s) → 4Fe(s) + 3CO2(g) What mass of Fe2O3 is necessary to produce 100. L of CO2 at 300. K and 2.10 atm? 823 g Fe2O3 908 g Fe2O3 1,110 g Fe2O3 1,360 g Fe2O3

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Answer:

908 g Fe₂O₃

General Formulas and Concepts:

Chemistry - Gas Laws

Combined Gas Law: PV = nRT

  • P is pressure
  • V is volume (in liters)
  • n is number of moles
  • R is a constant - [tex]0.0821 \frac{L \cdot atm}{mol \cdot K}[/tex]
  • T is temperature (in Kelvins)

Chemistry - Stoichiometry

  • Reading a Periodic Table
  • Using Dimensional Analysis

Explanation:

Step 1: Define

[RxN]   2Fe₂O₃ (s) + 3C (s) → 4Fe (s) + 3CO₂ (g)

[Given]   100 L CO₂ at 300 K and 2.10 atm

Step 2: Find Moles

Apply the Combined Gas Law and solve.

[tex](2.10 \ atm)(100 \ L) = n(0.0821 \frac{L \cdot atm}{mol \cdot K} )(300 \ K)\\\frac{(2.10 \ atm)(100 \ L)}{(0.0821 \frac{L \cdot atm}{mol \cdot K} )(300 \ K)} = n\\n = 8.52619 \ mol \ CO_2[/tex]

Step 3: Identify Conversions

[RxN]   3 mol CO₂ = 2 mol Fe₂O₃

Molar Mass of Fe - 55.85 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Fe₂O₃ - 2(55.85) + 3(16.00) = 159.7 g/mol

Step 4: Stoichiometry

[tex]8.52619 \ mol \ CO_2(\frac{2 \ mol \ Fe_2O_3}{3 \ mol \ CO_2} )(\frac{159.7 \ g \ Fe_2O_3}{1 \ mol \ Fe_2O_3} )[/tex] = 907.755 g Fe₂O₃

Step 5: Check

We are given 3 sig figs. Follow sig fig rules and round.

907.755 g Fe₂O₃ ≈ 908 g Fe₂O₃

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