The image illustrates the
Law of Conservation of Mass using a chemical reaction. Let's break it down step by step:
Chemical Reaction Shown
The reaction depicted is the combustion of methane (\( \text{CH}_4 \)) in the presence of oxygen (\( \text{O}_2 \)). The products are carbon dioxide (\( \text{CO}_2 \)) and water (\( \text{H}_2\text{O} \)).
The balanced chemical equation for this reaction is:
\[
\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}
\]
Molecular Representations
-
Methane (\( \text{CH}_4 \)): One molecule with one carbon atom (black) and four hydrogen atoms (yellow).
-
Oxygen (\( \text{O}_2 \)): Two molecules, each consisting of two oxygen atoms (cyan).
-
Carbon Dioxide (\( \text{CO}_2 \)): One molecule with one carbon atom (black) and two oxygen atoms (cyan).
-
Water (\( \text{H}_2\text{O} \)): Two molecules, each consisting of one oxygen atom (cyan) and two hydrogen atoms (yellow).
Mass Calculation
The Law of Conservation of Mass states that the total mass of the reactants must equal the total mass of the products. Let's calculate the masses:
####
Reactants:
1.
Methane (\( \text{CH}_4 \)):
- Molar mass of \( \text{CH}_4 \): \( 12 \, \text{g/mol} \) (carbon) + \( 4 \times 1 \, \text{g/mol} \) (hydrogen) = \( 16 \, \text{g/mol} \).
- Mass of \( \text{CH}_4 \): \( 16 \, \text{g} \).
2.
Oxygen (\( 2\text{O}_2 \)):
- Molar mass of \( \text{O}_2 \): \( 2 \times 16 \, \text{g/mol} \) (oxygen) = \( 32 \, \text{g/mol} \).
- Mass of \( 2\text{O}_2 \): \( 2 \times 32 \, \text{g} = 64 \, \text{g} \).
-
Total mass of reactants: \( 16 \, \text{g} + 64 \, \text{g} = 80 \, \text{g} \).
####
Products:
1.
Carbon Dioxide (\( \text{CO}_2 \)):
- Molar mass of \( \text{CO}_2 \): \( 12 \, \text{g/mol} \) (carbon) + \( 2 \times 16 \, \text{g/mol} \) (oxygen) = \( 44 \, \text{g/mol} \).
- Mass of \( \text{CO}_2 \): \( 44 \, \text{g} \).
2.
Water (\( 2\text{H}_2\text{O} \)):
- Molar mass of \( \text{H}_2\text{O} \): \( 2 \times 1 \, \text{g/mol} \) (hydrogen) + \( 16 \, \text{g/mol} \) (oxygen) = \( 18 \, \text{g/mol} \).
- Mass of \( 2\text{H}_2\text{O} \): \( 2 \times 18 \, \text{g} = 36 \, \text{g} \).
-
Total mass of products: \( 44 \, \text{g} + 36 \, \text{g} = 80 \, \text{g} \).
Verification of the Law of Conservation of Mass
-
Total mass of reactants: \( 80 \, \text{g} \)
-
Total mass of products: \( 80 \, \text{g} \)
Since the total mass of the reactants equals the total mass of the products, the Law of Conservation of Mass is satisfied.
Final Answer
\[
\boxed{80 \, \text{g}}
\]
Parent Tip: Review the logic above to help your child master the concept of law of conservation of mass example problems.
