Humboldt State University ® Department of Chemistry

Richard A. Paselk

	General Chemistry	Fall 2008
	© R. Paselk 2008
	Discussion Modules
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Reaction Stoichiometry

1. How many moles of carbon dioxide are produced in the complete combustion of 2.85 kg of methane (natural gas, CH₄)? How many grams?

For this example let's first write a balanced equation for the reaction. So what is going on?

• Combustion means burning, so the methane is reacting with oxygen, and of course atmospheric oxygen occurs as a diatomic molecule, O₂.
• The reaction side is therefore CH₄ + O₂
• Looking at the atoms on the reactant side and noting we have complete combustion we can guess that the products will be water (from hydrogen and oxygen) and carbon dioxide (from carbon and oxygen). This gives the unbalanced equation:

CH₄ + O₂ CO₂ + H₂O

By inspection we can see that carbon is already balanced. Balancing hydrogen then gives:

CH₄ + 2 O₂ CO₂ + 2 H₂O

Finally, balancing oxygen by inspection gives the complete balanced equation:

CH₄ + 2 O₂ CO₂ + 2 H₂O

So the reaction equation shows that there is a 1:1 ratio of CH₄: CO₂

Our next task is then to find how many moles we have of methane which is then the number of mole we can make of carbon dioxide. From the Periodic table we can read the atomic weights of carbon and hydrogen:

IA

IIA

1 H 1.008

IIIA

IVA

VA

VIA

VIIA

VIIIA

6 C 12.01

8 O 16.00

IIIB

IVB

VB

VI

VIIB

VIIIB

IB

IIB

• MW of methane = 1(12.01) + 4(1.008) = 16.04 g/mol
• Moles of methane = (2.85 kg CH₄)(1,000g/kg)/(16.04 g/mol) = 177.68 moles = 177 mol Carbon dioxide
• To find the grams we first need the MW of carbon dioxide = 12.01 + 2(16.00) = 44.04 g/mol
• Grams of carbon dioxide = (177.68 mol)(44.04 g/mol) = 7.8250x10³ g = 7.82x10³ g or 7.83x10³ g

2. How many kilograms of carbon dioxide are produced in the complete combustion of 42 kg of gasoline (assume gasoline is octane, C₈H₁₈)?

First balance the chemical equation by inspection:

C₈H₁₈ + 25/2 O₂ 8 CO₂ + 9 H₂O

2 C₈H₁₈ + 25 O₂ 16 CO₂ + 18 H₂O

From the equation, combustion of one mole of octane gives eight moles of carbon dioxide.

• The MW of octane = 8(12.01) + 18(1.008) = 114.22 g/mol
• Moles of octane = (42 kg)(1,000g/kg)/(114.22 g/mol) = 367.70 mol
• Moles of carbon dioxide = ( 367.70 mol octane)(8 mol carbon dioxide/1 mol octane) = 2,941.6
• Kilograms of carbon dioxide = (2,941.6 mol)(44.04 g CO₂/mol)(1 kg/1,000 g)= 1.2955x10² kg = 1.3x10² kg

3. How many grams of iron(III) oxide can be made from 22.0 g of iron?

Let's begin by finding the moles of iron = (22.0 g Fe)/(55.85 g/mol) = 0.3939 mol Fe

Next, writing and balancing the reaction equation we get:

2 Fe + 3/2 O₂ Fe₂O₃

Thus for moles of iron(III) Oxide we get (0.3939 mol Fe)(1 mol Fe₂O₃)/(2 mol Fe) = 0.19696 mol Fe₂O₃

Formula weight of Fe₂O₃ = 2(55.85) + 3(16.00) = 159.70 g/mol

Grams of iron(III) oxide = (0.19696 mol Fe₂O₃)(159.70 g/mol Fe₂O₃) = 31.455g = 31.5g

4. How many grams of oxygen will be used in the complete combustion of 50.67 g of propane (C₃H₈)?

Moles of propane = (50.67 g)/(3x12.01 g/mol + 8x1.008 g/mol) = 1.1491 mol

The balanced equation is:

C₃H₈ + 5 O₂ 3 CO₂ + 4 H₂O

Grams of oxygen = {(1.1491 mol C₃H₈)(5 mol O₂)/(mol C₃H₈)}{32.00 g O/mol O₂} = 183.86 g = 183.9 g

5. How many moles of oxygen are needed for the complete combustion of 45.9 g of methane?

Looking at the balanced equation:

CH₄ + 2 O₂ CO₂ + 2 H₂O

we see that we need two moles of oxygen (O₂) for each mole of methane

(2 mol oxygen/mol methane)(45.9 g methane)/(16.04 g methane/mol methane) = 5.7232 mol oxygen = 5.72 mol

© R A Paselk

Last modified 11 February 2013