Chemical equations are the language of chemistry, describing the changes that occur during a chemical reaction. These equations allow chemists to predict the products of a reaction, determine the amounts of reactants and products, and study the underlying chemical processes that occur.
Stoichiometry is the study of the quantitative relationships between reactants and products in a chemical reaction. By using stoichiometric calculations, chemists can determine the amount of reactants required to produce a certain amount of product, or the amount of product that can be produced from a given amount of reactants.
Let us explore the fundamentals of chemical equations and stoichiometry, including the principles behind balancing equations, mole-to-mole conversions, and limiting reactants.
Chemical Equations
A chemical equation is a representation of a chemical reaction using symbols and formulas. It describes the reactants, products, and the conditions under which the reaction occurs. The general form of a chemical equation is:
Reactants → Products
For example, the equation for the reaction between hydrogen gas and oxygen gas to form water is:
2H2 + O2 → 2H2O
The reactants are hydrogen gas (H2) and oxygen gas (O2), and the product is water (H2O). The coefficients in front of the formulas represent the number of moles of each compound involved in the reaction.
Balancing Chemical Equations
The Law of Conservation of Mass states that the mass of the reactants must be equal to the mass of the products in a chemical reaction. This means that the number of atoms of each element on the reactant side must be the same as the number of atoms of each element on the product side.
To balance a chemical equation, we adjust the coefficients in front of the formulas to ensure that the number of atoms of each element is equal on both sides of the equation. For example, let’s balance the equation for the combustion of propane (C3H8) in air to form carbon dioxide and water:
C3H8 + O2 → CO2 + H2O
To balance this equation, we start by counting the number of atoms of each element on each side of the equation:
Reactants: C – 3, H – 8, O – 2 Products: C – 1, H – 2, O – 3
To balance the carbons, we add a coefficient of 3 in front of the CO2:
C3H8 + O2 → 3CO2 + H2O
Now we have 3 carbon atoms on each side. To balance the hydrogens, we add a coefficient of 4 in front of the H2O:
C3H8 + O2 → 3CO2 + 4H2O
Now we have 8 hydrogen atoms on each side. Finally, we balance the oxygens by adding a coefficient of 5/2 in front of the O2:
C3H8 + 5/2 O2 → 3CO2 + 4H2O
Now we have 10 oxygen atoms on each side, and the equation is balanced.
Related Post: Understanding Atoms and Molecules
Stoichiometry
Stoichiometry is the study of the quantitative relationships between reactants and products in a chemical reaction. The principles of stoichiometry can be used to determine the amount of reactants required to produce a certain amount of product, or the amount of product that can be produced from a given amount of reactants.
Mole-to-Mole Conversions
In stoichiometric calculations, we use the mole ratio between reactants and products to determine the amounts of each substance involved in a reaction. The mole ratio is determined by the coefficients in the balanced chemical equation.
For example, let’s consider the reaction between hydrogen gas and nitrogen gas to form ammonia:
N2 + 3H2 → 2NH3
The mole ratio between nitrogen gas and ammonia is 1:2, which means that for every 1 mole of nitrogen gas that reacts, 2 moles of ammonia are produced. Similarly, the mole ratio between hydrogen gas and ammonia is 3:2, which means that for every 3 moles of hydrogen gas that reacts, 2 moles of ammonia are produced.
To determine the amount of ammonia that can be produced from a given amount of nitrogen gas, we use the mole ratio between nitrogen gas and ammonia. For example, if we have 2 moles of nitrogen gas, we can calculate the amount of ammonia produced as follows:
2 moles N2 x (2 moles NH3/1 mole N2) = 4 moles NH3
Similarly, to determine the amount of hydrogen gas required to produce a certain amount of ammonia, we use the mole ratio between hydrogen gas and ammonia. For example, if we want to produce 6 moles of ammonia, we can calculate the amount of hydrogen gas required as follows:
6 moles NH3 x (3 moles H2/2 moles NH3) = 9 moles H2
Limiting Reactants
In a chemical reaction, the limiting reactant is the reactant that is completely consumed and limits the amount of product that can be formed. The excess reactant is the reactant that is not completely consumed and remains in the reaction mixture after the reaction is complete.
To determine the limiting reactant and the amount of product that can be formed, we must compare the amounts of each reactant present to the mole ratio in the balanced chemical equation. The reactant that produces the least amount of product is the limiting reactant.
For example, consider the reaction between 2 moles of hydrogen gas and 1 mole of oxygen gas to form water:
2H2 + O2 → H2O
The balanced chemical equation tells us that 2 moles of hydrogen gas react with 1 mole of oxygen gas to form 2 moles of water. If we have 4 moles of hydrogen gas and 2 moles of oxygen gas, we can determine the limiting reactant as follows:
Hydrogen gas: 4 moles H2 x (1 mole O2/2 moles H2) = 2 moles O2 required Oxygen gas: 2 moles O2
Since we need 2 moles of oxygen gas to react with 4 moles of hydrogen gas, the oxygen gas is the limiting reactant. The maximum amount of water that can be formed is determined by the amount of limiting reactant, which is 2 moles of oxygen gas. The excess reactant is the hydrogen gas, which will not be completely consumed and will remain in the reaction mixture after the reaction is complete.
Percent Yield
In a chemical reaction, the theoretical yield is the amount of product that would be obtained if all of the limiting reactant were consumed and reacted completely to form the product. The actual yield is the amount of product that is actually obtained from the reaction. The percent yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage.
Percent yield = (actual yield/theoretical yield) x 100%
The percent yield can be used to determine the efficiency of a chemical reaction and to identify sources of error or inefficiency in the reaction.
For example, if the theoretical yield of a reaction is 10 grams and the actual yield is 8 grams, the percent yield is:
Percent yield = (8/10) x 100% = 80%
This means that the reaction was 80% efficient and that there may have been some sources of error or inefficiency in the reaction.
Exercise 1
If 2 moles of aluminum react with 3 moles of chlorine gas, what is the theoretical yield of aluminum chloride?
Balanced equation: 2Al + 3Cl2 → 2AlCl3
2 moles Al x (3 moles Cl2/2 moles Al) = 3 moles Cl2 required 3 moles Cl2
Since we have enough chlorine gas to react with all of the aluminum, the limiting reactant is aluminum.
Therefore, the theoretical yield of aluminum chloride is 2 moles.
Exercise 2
A reaction between 25.0 grams of magnesium and 50.0 grams of oxygen gas produces magnesium oxide. What is the limiting reactant and the theoretical yield of magnesium oxide?
Balanced equation: 2Mg + O2 → 2MgO, molar mass of Mg = 24.31 g/mol,
molar mass of O2 = 32.00 g/mol
To determine the limiting reactant and theoretical yield, we first need to convert the masses of reactants to moles:
25.0 g Mg x (1 mol Mg/24.31 g) = 1.03 mol Mg 50.0 g O2 x (1 mol O2/32.00 g)
The above balanced equation shows that 2 moles of aluminum react with 3 moles of chlorine gas to form 2 moles of aluminum chloride. Therefore, the theoretical yield of aluminum chloride is directly proportional to the amount of limiting reactant. To determine the limiting reactant, we need to compare the amount of each reactant to the stoichiometric ratio:
= 1.56 mol O2
Next, we need to compare the amount of each reactant to the stoichiometric ratio:
1.03 mol Mg x (1 mol O2/2 mol Mg) = 0.515 mol O2 required 1.56 mol O2
Since we have more oxygen gas than required, the limiting reactant is magnesium. The theoretical yield of magnesium oxide can be calculated using the stoichiometric ratio:
1.03 mol Mg x (2 mol MgO/2 mol Mg) x (40.31 g MgO/1 mol MgO) = 83.1 g MgO
Therefore, the theoretical yield of magnesium oxide is 83.1 grams.
In conclusion, chemical equations and stoichiometry are important concepts in the study of chemistry. Chemical equations describe the reactants and products involved in a chemical reaction, while stoichiometry involves the quantitative relationships between these reactants and products. By understanding these concepts, we can predict and control chemical reactions and determine the amount of product that can be obtained from a given amount of reactants.
