The image shows a chemical equation and the calculation of molar masses for the reactants and products involved in the reaction. Let's break this down step by step:
Chemical Equation
The given chemical equation is:
\[
6O_2 + C_6H_{12}O_6 \rightarrow 6CO_2 + 6H_2O
\]
This represents the process of cellular respiration, where glucose (\(C_6H_{12}O_6\)) reacts with oxygen (\(O_2\)) to produce carbon dioxide (\(CO_2\)) and water (\(H_2O\)).
Molar Mass Calculations
####
1. Molar Mass of \(O_2\):
- The molecular formula of oxygen gas is \(O_2\).
- Each oxygen atom has an atomic mass of approximately \(16 \, \text{g/mol}\).
- Therefore, the molar mass of \(O_2\) is:
\[
2 \times 16 \, \text{g/mol} = 32 \, \text{g/mol}
\]
####
2. Molar Mass of \(C_6H_{12}O_6\):
- The molecular formula of glucose is \(C_6H_{12}O_6\).
- To calculate its molar mass, we sum the contributions from each element:
- Carbon (\(C\)): There are 6 carbon atoms, and the atomic mass of carbon is \(12 \, \text{g/mol}\):
\[
6 \times 12 \, \text{g/mol} = 72 \, \text{g/mol}
\]
- Hydrogen (\(H\)): There are 12 hydrogen atoms, and the atomic mass of hydrogen is \(1 \, \text{g/mol}\):
\[
12 \times 1 \, \text{g/mol} = 12 \, \text{g/mol}
\]
- Oxygen (\(O\)): There are 6 oxygen atoms, and the atomic mass of oxygen is \(16 \, \text{g/mol}\):
\[
6 \times 16 \, \text{g/mol} = 96 \, \text{g/mol}
\]
- Adding these together gives the total molar mass of glucose:
\[
72 \, \text{g/mol} + 12 \, \text{g/mol} + 96 \, \text{g/mol} = 180 \, \text{g/mol}
\]
Summary of Molar Masses
- Molar mass of \(O_2\): \(32 \, \text{g/mol}\)
- Molar mass of \(C_6H_{12}O_6\): \(180 \, \text{g/mol}\)
Explanation
The problem involves calculating the molar masses of the reactants in the given chemical equation. This is a fundamental step in stoichiometry, which helps in understanding the quantitative relationships between reactants and products in a chemical reaction.
Final Answer
\[
\boxed{180 \, \text{g/mol}}
\] (for \(C_6H_{12}O_6\))
Parent Tip: Review the logic above to help your child master the concept of percent yield equation.
