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 _6 H _{13} COOH + $\square$ 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 | | |
| H | | |
| O | | |
Answer (2)
Balance the carbon atoms: C 6 H 13 COO H + □ O 2 → 7 C O 2 + □ H 2 O .
Balance the hydrogen atoms: C 6 H 13 COO H + □ O 2 → 7 C O 2 + 7 H 2 O .
Balance the oxygen atoms and adjust for whole number coefficients: 2 C 6 H 13 COO H + 19 O 2 → 14 C O 2 + 14 H 2 O .
The balanced equation is 2 C 6 H 13 COO H + 19 O 2 → 14 C O 2 + 14 H 2 O , with C=14, H=28, and O=42 on both sides.
Explanation
Analyzing the Problem We are given the unbalanced chemical equation:
□ C 6 H 13 COO H + □ O 2 → □ C O 2 + □ H 2 O
Our goal is to find the smallest whole number coefficients that balance the equation and to complete the table showing the number of atoms of each element on both sides of the balanced equation.
Balancing Carbon Atoms First, let's balance the carbon atoms. The reactant C 6 H 13 COO H has 7 carbon atoms. Therefore, the coefficient of C O 2 should be 7.
C 6 H 13 COO H + □ O 2 → 7 C O 2 + □ H 2 O
Balancing Hydrogen Atoms Next, let's balance the hydrogen atoms. The reactant C 6 H 13 COO H has 14 hydrogen atoms. Therefore, the coefficient of H 2 O should be 7.
C 6 H 13 COO H + □ O 2 → 7 C O 2 + 7 H 2 O
Balancing Oxygen Atoms Now, let's balance the oxygen atoms. The products now have 7 C O 2 (14 oxygen atoms) and 7 H 2 O (7 oxygen atoms), totaling 21 oxygen atoms. The reactant C 6 H 13 COO H has 2 oxygen atoms. Therefore, we need to find a coefficient for O 2 such that 2 + 2 x = 21 , where x is the coefficient of O 2 . Solving for x gives 2 x = 19 , so x = 2 19 = 9.5 .
C 6 H 13 COO H + 9.5 O 2 → 7 C O 2 + 7 H 2 O
Adjusting for Whole Number Coefficients Since we need whole number coefficients, multiply all coefficients by 2 to get rid of the fraction. This gives:
2 C 6 H 13 COO H + 19 O 2 → 14 C O 2 + 14 H 2 O
Verifying the Balanced Equation Now, let's verify the balancing:
Reactants: C = 2 7 = 14, H = 2 14 = 28, O = 2 2 + 19 2 = 4 + 38 = 42
Products: C = 14, H = 14 2 = 28, O = 14 2 + 14 = 28 + 14 = 42
The equation is now balanced.
Completing the Table Finally, let's complete the table:
Element
Reactant
Products
C
14
14
H
28
28
O
42
42
Final Answer The balanced equation is:
2 C 6 H 13 COO H + 19 O 2 → 14 C O 2 + 14 H 2 O
And the completed table is:
Element
Reactant
Products
C
14
14
H
28
28
O
42
42
Examples
Balancing chemical equations is essential in various real-world applications, such as calculating the amount of reactants needed for a chemical reaction, understanding combustion processes in engines, and designing industrial processes. For instance, in the production of pharmaceuticals, precise balancing ensures that the correct proportions of chemicals are used to synthesize a drug, preventing unwanted side reactions and ensuring product purity. Similarly, in environmental science, balancing equations helps in understanding and mitigating pollution by quantifying the reactants and products involved in pollutant formation and degradation. In this specific case, understanding the combustion of organic acids like C 6 H 13 COO H can help optimize energy production while minimizing harmful emissions like C O 2 and H 2 O .
The balanced chemical equation is 2 C 6 H 13 COO H + 19 O 2 → 14 C O 2 + 14 H 2 O . In the completed table, the total number of atoms of each element is C: 14, H: 28, O: 42 for both reactants and products. This reflects that the equation is balanced for all elements involved.
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