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Chapter 8: Problem 28

Balance this chemical equation by inspection: \(\mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\)

Short Answer

Expert verified
The balanced chemical equation is: \[\mathrm{C}_{2} \mathrm{H}_{4}+3\,\mathrm{O}_{2} \rightarrow 2\,\mathrm{CO}_{2}+2\,\mathrm{H}_{2} \mathrm{O}\]

Step by step solution

01

Identify the elements and their initial counts

On the left side of the equation, there are 2 carbon atoms, 4 hydrogen atoms, and 2 oxygen atoms. On the right side, there is 1 carbon atom, 2 hydrogen atoms, and 3 oxygen atoms.
02

Balance the carbon atoms

To balance the carbon atoms, we need an equal number of carbon atoms on both sides. Since there are 2 carbon atoms on the left side and only 1 on the right, we need to multiply the carbon dioxide (CO2) term by 2. This will give us the balanced equation: \[\mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{O}_{2} \rightarrow 2\,\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\]
03

Balance the hydrogen atoms

Now, let's balance the hydrogen atoms. There are 4 hydrogen atoms on the left side and 2 on the right. We need to multiply the water (H2O) term by 2 to get the same number of hydrogen atoms on both sides. This gives us the balanced equation: \[\mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{O}_{2} \rightarrow 2\,\mathrm{CO}_{2}+2\,\mathrm{H}_{2} \mathrm{O}\]
04

Balance the oxygen atoms

Lastly, we have to balance the oxygen atoms. After the previous steps, there are 2 oxygen atoms on the left side (in O2) and 6 oxygen atoms on the right side (4 in the two CO2 and 2 in the two H2O). To balance the oxygens, we need to multiply the oxygen gas (O2) term by 3. This gives us the final balanced equation: \[\mathrm{C}_{2} \mathrm{H}_{4}+3\,\mathrm{O}_{2} \rightarrow 2\,\mathrm{CO}_{2}+2\,\mathrm{H}_{2} \mathrm{O}\] The chemical equation is now balanced with 2 carbon atoms, 4 hydrogen atoms, and 6 oxygen atoms on both sides.

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Most popular questions from this chapter

Balance this chemical equation and assign it a reaction type: \(\mathrm{Li}(\mathrm{s})+\mathrm{N}_{2}(g) \rightarrow \mathrm{Li}_{3} \mathrm{~N}(s)\)

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