Fill in the gaps to balance the equation.
Use the smallest set of whole numbers to balance the equation and include coefficients of "1" when appropriate.
[tex]$\square$ C _2 H _6+ $\square$ O _2 \rightarrow $\square$ CO _2+ $\square$ H _2 O[/tex]
Complete the table to determine how many atoms of each element are present in the reactants and products.
| Element | Reactant | Products |
| :------ | :------- | :------- |
| C | $\square$ | $\square$ |
| H | $\square$ | $\square$ |
| O | $\square$ | $\square$ |
Answer (1)
Balance the carbon atoms by placing a coefficient of 2 in front of C O 2 : C 2 H 6 + O 2 → 2 C O 2 + H 2 O .
Balance the hydrogen atoms by placing a coefficient of 3 in front of H 2 O : C 2 H 6 + O 2 → 2 C O 2 + 3 H 2 O .
Balance the oxygen atoms by placing a coefficient of 2 7 in front of O 2 and then multiply all coefficients by 2 to get whole numbers: 2 C 2 H 6 + 7 O 2 → 4 C O 2 + 6 H 2 O .
Complete the table with the number of atoms of each element on both sides. The balanced equation is 2 C 2 H 6 + 7 O 2 → 4 C O 2 + 6 H 2 O . The final answer is:
Element C H O Reactant 4 12 14 Products 4 12 14
2 C 2 H 6 + 7 O 2 → 4 C O 2 + 6 H 2 O
Explanation
Initial Analysis and Strategy Let's balance the chemical equation for the combustion of ethane ( C 2 H 6 ). Our goal is to find the smallest whole number coefficients that satisfy the conservation of mass for each element. The unbalanced equation is:
C 2 H 6 + O 2 → C O 2 + H 2 O
We will balance the equation step by step, starting with carbon and hydrogen, and then oxygen.
Balancing Carbon First, let's balance the carbon atoms. There are 2 carbon atoms on the left side ( C 2 H 6 ), so we need 2 carbon atoms on the right side. We place a coefficient of 2 in front of C O 2 :
C 2 H 6 + O 2 → 2 C O 2 + H 2 O
Now, the carbon atoms are balanced.
Balancing Hydrogen Next, we balance the hydrogen atoms. There are 6 hydrogen atoms on the left side ( C 2 H 6 ), so we need 6 hydrogen atoms on the right side. We place a coefficient of 3 in front of H 2 O :
C 2 H 6 + O 2 → 2 C O 2 + 3 H 2 O
Now, the hydrogen atoms are balanced.
Balancing Oxygen and Achieving Whole Number Coefficients Now, let's balance the oxygen atoms. On the right side, we have 2 × 2 = 4 oxygen atoms from C O 2 and 3 × 1 = 3 oxygen atoms from H 2 O , for a total of 4 + 3 = 7 oxygen atoms. We need 7 oxygen atoms on the left side. We can place a coefficient of 2 7 in front of O 2 :
C 2 H 6 + 2 7 O 2 → 2 C O 2 + 3 H 2 O
However, we need whole number coefficients. To achieve this, we multiply all coefficients by 2:
2 C 2 H 6 + 7 O 2 → 4 C O 2 + 6 H 2 O
Now, the equation is balanced with whole number coefficients.
Completing the Atom Count Table Finally, we complete the table to show the number of atoms of each element on both sides of the balanced equation:
Element
Reactant
Products
C
2 × 2 = 4
4 × 1 = 4
H
2 × 6 = 12
6 × 2 = 12
O
7 × 2 = 14
( 4 × 2 ) + ( 6 × 1 ) = 8 + 6 = 14
The number of atoms of each element is the same on both sides, confirming that the equation is balanced.
Final Balanced Equation and Table The balanced equation is:
2 C 2 H 6 + 7 O 2 → 4 C O 2 + 6 H 2 O
The completed table is:
Element
Reactant
Products
C
4
4
H
12
12
O
14
14
Examples
Balancing chemical equations is essential in many real-world applications, such as calculating the amount of reactants needed for a chemical reaction or determining the amount of products formed. For example, in the combustion of fuels like ethane, it's crucial to know the exact amount of oxygen required to ensure complete combustion and minimize the production of harmful byproducts like carbon monoxide. In industrial processes, balanced equations help optimize reactions, reduce waste, and improve efficiency. Understanding stoichiometry through balanced equations is fundamental in fields like environmental science, chemical engineering, and materials science.
For instance, if you want to burn 100 grams of ethane ( C 2 H 6 ), you would need to calculate the exact amount of oxygen required based on the balanced equation 2 C 2 H 6 + 7 O 2 → 4 C O 2 + 6 H 2 O . This ensures that the combustion is complete and efficient, maximizing energy output and minimizing pollution.
