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C4 H16 + 02 CO2 + H2O Balance the equation and identify the reaction type Answer choices may be used more than once.

C4 H16 + 02 CO2 + H2O Balance the equation and identify the reaction type Answer choices may be used more than once.

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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### Balancing Chemical Equations: Combustion Reaction **Equation to Balance:** \[ \text{C}_4\text{H}_{16} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \] **Task:** Balance the chemical equation and identify the type of chemical reaction. *Answer choices may be used more than once.* **Dropdown Selections:** - **Oxygen:** 2 - **Water:** 8 - **Carbon Dioxide:** 2 - **Hydrocarbon:** [Choose] - **Reaction Type:** Combustion **Explanation of Reaction Type:** This is a combustion reaction because it involves the burning of a hydrocarbon (\(\text{C}_4\text{H}_{16}\)) in oxygen to produce carbon dioxide and water. **Instructions:** Carefully adjust the coefficients to ensure that both sides of the equation have an equal number of each type of atom.
Transcribed Image Text:### Balancing Chemical Equations: Combustion Reaction **Equation to Balance:** \[ \text{C}_4\text{H}_{16} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \] **Task:** Balance the chemical equation and identify the type of chemical reaction. *Answer choices may be used more than once.* **Dropdown Selections:** - **Oxygen:** 2 - **Water:** 8 - **Carbon Dioxide:** 2 - **Hydrocarbon:** [Choose] - **Reaction Type:** Combustion **Explanation of Reaction Type:** This is a combustion reaction because it involves the burning of a hydrocarbon (\(\text{C}_4\text{H}_{16}\)) in oxygen to produce carbon dioxide and water. **Instructions:** Carefully adjust the coefficients to ensure that both sides of the equation have an equal number of each type of atom.
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