Enduring Understanding 3.A.2: Stoichiometry

The ratios of coefficients from balanced equations can be used to determine the amounts of substances (mass of substance, volume of gas or solution) that are consumed or produced in a chemical reaction.

Amount of reactant <--> moles of reactant <--> moles of product <--> amount of product

Example: When 4.0 grams of hydrogen react with excess oxygen, what mass of water is produced?

As seen before, the balanced equation is: 2H₂ + O₂ → 2H₂O

The coefficient for hydrogen is 2, and for water is 2, so the same number of moles of water will be produced as of hydrogen that reacts.

# moles hydrogen reacting = 4.0 g / 2.02 g/mol = 2.0 mol

Therefore, # moles water formed = 2.0 mol

Mass of water formed = 2.0 mol x 18.02 g/mol = 36 g

Chemical reaction problems will often be presented as limiting reagent problems. Because atoms and molecules react in definite and fixed proportions, sometimes there will be too much of one or more reagents for that reagent(s) to be completely consumed.
In such problem, the relative molar amounts of the reactants and the stoichiometry of the reaction must be used to determine which reagent is limiting and which is/are in excess.

Example: Given the redox reaction: MnO₄^- + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

How many moles of Mn²⁺ are formed if 2 moles MnO₄^- and 2.5 moles Fe²⁺ are reacted together?

This is a limiting reagent question. Since 1 mole of MnO₄^- reacts with 5 moles of Fe²⁺:

2 moles of MnO₄^- would react with 10 moles of Fe²⁺, or...

0.5 moles of MnO₄^- would react with 2.5 moles of Fe²⁺

Therefore, Fe²⁺ is the limiting reagent. For every 5 moles Fe²⁺, 1 mole of Mn²⁺ is formed.

2.5 moles of Fe²⁺ reacts, therefore 0.5 moles of Mn²⁺ is formed.

Percent yield is the percentage of the theoretical amount of a product actually isolated from a reaction.

Example: In the above MnO₄^-/Fe²⁺ redox reaction, suppose 0.4 moles of Mn²⁺ was isolated as a salt. The percent yield in this case would be 0.4 moles/0.5 moles = 80 %.

Sample Question: Calculate the mass of oxygen required to react with 20.0 g sulfur to yield sulfur dioxide, according to the equation:

S₈ + 8 O₂ → 8 SO₂

First, calculate the molar amount of sulfur reacting:

20.0 g/ 256.52 g/mol = 0.0780 mol S₈

The molar ratio of oxygen to sulphur in the balanced equation is 8:1, so the molar amount of oxygen reacting is:

8/1 x 0.0780 mol = 0.624 mol O₂

Now determine the mass of oxygen required:

0.624 mol x 32.00 g/mol = 20.0 g O₂

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Enduring Understanding 3.A.2: Stoichiometry

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