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Chapter 9: Problem 114

(a) Write a balanced chemical equation for the combustion of \(\mathrm{CH}_{4}(\mathrm{~g})\) with \(\mathrm{O}_{2}(g)\) to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\) (b) Consider the reaction of three molecules of \(\mathrm{CH}_{4}(g)\) with four molecules of \(\mathrm{O}_{2}(g)\). The following boxes represent before and after pictures of the tiny, sealed container in which this reaction takes place. Draw pictures to show the contents of this container before and after the reaction happens:

Short Answer

Expert verified
(a) The balanced chemical equation for the combustion of CH4(g) with O2(g) to form CO2(g) and H2O(g) is: \[\mathrm{CH}_{4}(\mathrm{g}) + 2\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}) + 2\mathrm{H}_{2}\mathrm{O}(\mathrm{g})\] (b) The initial container has three molecules of CH4(g) and four molecules of O2(g). After the reaction, the container will have 0 molecules of CH4(g), 1 molecule of O2(g), 3 molecules of CO2(g), and 3 molecules of H2O(g). Draw pictures to represent the contents of the container before and after the reaction accordingly.

Step by step solution

01

Write the unbalanced chemical equation

Write the equation showing reactants and products: \[\mathrm{CH}_{4}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2}\mathrm{O}(\mathrm{g})\]
02

Balance the chemical equation

To balance the equation, we need to adjust the coefficients in front of each compound so that the number of atoms of each element is the same on both sides of the equation. \[\mathrm{CH}_{4}(\mathrm{g}) + 2\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}) + 2\mathrm{H}_{2}\mathrm{O}(\mathrm{g})\] Now we have a balanced chemical equation for the combustion of CH4(g) with O2(g) to form CO2(g) and H2O(g). For part (b):
03

Draw the initial state of the container

Based on the given condition, the container has three molecules of CH4(g) and four molecules of O2(g) at the beginning. Draw a picture of the container with these molecules.
04

Calculate the number of molecules after the reaction

Using the balanced chemical equation, we can determine the number of molecules of each compound after the reaction. The reaction is: \[\mathrm{CH}_{4}(\mathrm{g}) + 2\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}) + 2\mathrm{H}_{2}\mathrm{O}(\mathrm{g})\] Given three molecules of CH4(g) and four molecules of O2(g), we can see that CH4 is the limiting reactant, as it would require six molecules of O2 to react with all the CH4 present. Therefore, we will only have 1.5 times the reaction happening. After the reaction, we will have: - 0 molecules of CH4(g) - 1 molecule of O2(g) (4 - 3*2/2 = 1) - 3 molecules of CO2(g) (3*1 = 3) - 3 molecules of H2O(g) (3*2/2 = 3)
05

Draw the final state of the container

Using the calculated numbers, draw a picture of the container with the products of the reaction and any remaining reactants. The container will now have 0 CH4 molecules, 1 O2 molecule, 3 CO2 molecules, and 3 H2O molecules.

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