Introduction
Reaction stoichiometry establishes the quantities of reactants (used) and products (obtained) based on a balanced chemical equation.
Chemical Reactions
A chemical change or chemical reaction can be described by writing a chemical equation. A chemical equation uses chemical symbols to show what happens during a chemical reaction.
Let's look at the electrolysis of water, a decomposition reaction. When water is subjected to an electric current, hydrogen gas and oxygen gas are formed. A chemical equation can be written to show this reaction:
H2O → H2 + O2
The arrow (→) means yields. The reactants are a substance that undergoes a change in a chemical reaction (the left side of the arrow). The products are a substance that is formed as a result of a chemical reaction (the right side of the arrow). The chemical equation is like a mathematical equation where both sides must be equal. Recall that mass cannot be created or destroyed. Therefore, we must balance the equation. Balancing equations is a "trial-and-error" method of equalizing the elements and molecules on both sides of the equation. When balancing an equation, the following guidelines can help:
- Write the unbalanced equation, including the correct formulas, for all reactants and products.
- Compare the number of atoms on the reactants and product(s) sides.
- Balance the elements by changing the number of molecules or ions with coefficients. Do not change the molecules or ions. The coefficients represent the number of moles of a substance. Always balance the heavier atoms before trying to balance lighter ones such as hydrogen.
- If necessary, continue to rebalance and recheck. Consider reducing the coefficients so that the smallest possible whole numbers are used. Fractions can be used for oxygen gas.
For the electrolysis of water, there are 2 hydrogen and 1 oxygen on the reactant side and 2 hydrogen and 2 oxygen on the product side of the equation.
A 2 can be added in front of the water to balance the oxygen:
However, this change unbalances the hydrogen, so a 2 can be added before the hydrogen gas in the product:
Because the electrolysis of water contains oxygen gas and fractions can be used, the equation could also be balanced:
H2O → H2 +  O2 |
Example:
Write the balanced equation for the combustion of methane and oxygen gas to yield carbon dioxide and water.
| 1. |
Write the unbalanced chemical equation: |
| |
CH4 + O2 → CO2 + H2O |
| 2. |
Identify the number of atoms: |
| |
CH4 + O2 → CO2 + H2O |
| 3. |
Balance the oxygen: |
| |
CH4 + 2O2 → CO2 + 2H2O |
| 4. |
Recheck (notice that the hydrogen is automatically balanced in this example): |
| |
Answer: CH4 + 2O2 → CO2 + 2H2O |
Reaction Stoichiometry
Stoichiometry establishes the quantities of reactants (used) and products (obtained) based on a balanced chemical equation. With a balanced equation, you can compare reactants and products, and determine the amount of products that might be formed or the amount or reactants needed to produce a certain amount of a product. However, when comparing different compounds in a reaction, you must always compare in moles (i.e., the coefficients). The different types of stoichiometric calculations are summarized in Figure 5.1.
Example:
How many grams of water are produced when 24.3 g of methane is reacted with excess oxygen?
| 1. |
Write the unbalanced equation: |
| |
CH4 + O2 → CO2 + H2O |
| 2. |
Balance the equation: |
| |
CH4 + 2O2 → CO2 + 2H2O |
| 3. |
Identify the amounts given and needed: |
| |
CH4 + 2O2 → CO2 + 2H2O |
| |
24.3 g → ? |
| 4. |
Use the molar masses and mole ratios (coeffi-cients) to set up the calculation: |
| |
 |
| |
 = 54.6 g H2O |
| |
Glucose (C6H12O6) ferments over time to produce ethanol (C2H5OH) and carbon dioxide. How many grams of glucose are needed to produce 100.0 g of ethanol? |
| 5. |
Write the unbalanced equation: |
| |
C6H12O6 → C2H5OH + CO2 |
| 6. |
Balance the equation: |
| |
C6H12O6 → 2C2H5OH + 2CO2 |
| 7. |
Identify the amounts given and needed: |
| |
C6H12O6 → 2C2H5OH + 2CO2 |
| |
? → 100.0 g |
| 8. |
Use the molar masses and mole ratios (coeffi-cients) to set up the calculation: |
| |
 |
Limiting Reactant
The reagent that is consumed first in a reaction is called the limiting reactant or reagent. In previous examples and problems, the assumption was that one reactant was in excess and the other was the limiting reactant. However, if a known amount of each reactant is added to a reaction vessel, then the limiting reactant must be calculated. Because the limiting reactant is consumed and limits the amount of products being formed, the easiest method of finding the limiting reactant is to simply calculate the amount of product that could be formed from each reactant. The reactant that produces the least amount of product is the limiting reagent. The amount of product that the limiting reactant can produce is called the theoretical yield of the reaction.
Example:
Identify the limiting reactant and how much ammonia gas can be produced when 8.0 g of nitrogen gas reacts with 8.0 g of hydrogen gas by the use of the Haber process: 3H2 + N2→ 2NH3.
Solution:
| 1. |
Identify the amount given and needed: |
| |
3H2 + N2 → 2NH3 |
| |
8.0 g + 8.0 g → ? |
| 2. |
Use the molar masses and mole ratios (coeffi-cients) to set up the calculation: |
| |
 |
| |
 |
| 3. |
The limiting reactant is nitrogen because it can produce the smallest amount of ammonia. The theoretical yield of ammonia is 9.7 g. |
Percent Yield
The percentage yield is a ratio of the actual yield of a product over the expected one, known as the theoretical yield.
Example:
From the previous example, what is the percent yield if only 8.2 g of ammonia is produced?
Solution:
Practice problems for these concepts can be found at - Stoichiometry and Chemical Equations Practice Questions
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From Chemistry Success in 20 Mintues A Day. Copyright © 2005 by LearningExpress, LLC. All Rights Reserved.
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