- Determine the moles of reactants and products using given molar masses and the stoichiometric ratios from the equation: \(6 \text{CO}_2 + 6 \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6 \text{O}_2\).
- Calculate the mass of water required for \(157.35 \, \text{g}\) of \(\text{CO}_2\) by converting to moles (\(\approx 3.5753 \, \text{mol}\)) and using the \(1:1\) molar ratio, resulting in \(\approx 64.43 \, \text{g}\) of \(\text{H}_2\text{O}\).
- Calculate the mass of oxygen formed from \(15.90 \, \text{g}\) of glucose by converting to moles (\(\approx 0.08824 \, \text{mol}\)) and using the \(1:6\) molar ratio, resulting in \(\approx 16.94 \, \text{g}\) of \(\text{O}_2\).
- The final results are \(157.35 \, \text{g} \, \text{CO}_2 = \boxed{64.43} \, \text{g} \, \text{H}_2\text{O}\) and \(15.90 \, \text{g} \, \text{C}_6\text{H}_{12}\text{O}_6 = \boxed{16.94} \, \text{g} \, \text{O}_2\).
Explanation
- Analysis of the Photosynthesis Reaction
The chemical equation for photosynthesis is:
\[6 \text{CO}_2 + 6 \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6 \text{O}_2\]
This equation shows that 6 moles of carbon dioxide \((CO_2)\) react with 6 moles of water \((H_2O)\) to produce 1 mole of glucose \((C_6H_{12}O_6)\) and 6 moles of oxygen \((O_2)\). Therefore, the molar ratio of \(\text{CO}_2\) to \(\text{H}_2\text{O}\) is \(6:6\), which simplifies to \(1:1\). The molar ratio of glucose \((C_6H_{12}O_6)\) to oxygen \((O_2)\) is \(1:6\).
- Calculating the Mass of Water Required
First, we determine the amount of water required to react with \(157.35 \, \text{g}\) of \(\text{CO}_2\).
- Convert the mass of \(\text{CO}_2\) to moles:
\[\text{Moles of } \text{CO}_2 = \frac{157.35 \, \text{g}}{44.01 \, \text{g/mol}} \approx 3.5753 \, \text{mol}\]
- Using the \(1:1\) molar ratio, the moles of \(\text{H}_2\text{O}\) required is also \(3.5753 \, \text{mol}\).
- Calculate the mass of \(\text{H}_2\text{O}\):
\[\text{Mass of } \text{H}_2\text{O} = 3.5753 \, \text{mol} \times 18.02 \, \text{g/mol} \approx 64.43 \, \text{g}\]
- Calculating the Mass of Oxygen Produced
Next, we determine the mass of oxygen \((O_2)\) formed from the production of \(15.90 \, \text{g}\) of glucose \((C_6H_{12}O_6)\).
- Convert the mass of \(\text{C}_6\text{H}_{12}\text{O}_6\) to moles:
\[\text{Moles of } \text{C}_6\text{H}_{12}\text{O}_6 = \frac{15.90 \, \text{g}}{180.18 \, \text{g/mol}} \approx 0.08824 \, \text{mol}\]
- Using the \(1:6\) molar ratio, the moles of \(\text{O}_2\) produced is:
\[\text{Moles of } \text{O}_2 = 6 \times 0.08824 \, \text{mol} \approx 0.5294 \, \text{mol}\]
- Calculate the mass of \(\text{O}_2\):
\[\text{Mass of } \text{O}_2 = 0.5294 \, \text{mol} \times 32.00 \, \text{g/mol} \approx 16.94 \, \text{g}\]
- Conclusion
Based on the calculations above, the mass of water required is approximately \(64.43 \, \text{g}\), and the mass of oxygen produced is approximately \(16.94 \, \text{g}\).
\[157.35 \, \text{g} \, \text{CO}_2 = \boxed{64.43} \, \text{g} \, \text{H}_2\text{O}\]
\[15.90 \, \text{g} \, \text{C}_6\text{H}_{12}\text{O}_6 = \boxed{16.94} \, \text{g} \, \text{O}_2\]
Examples
In environmental science, understanding these stoichiometric relationships is crucial for calculating the carbon sequestration potential of forests or for industrial processes like biomass energy production. By knowing the exact mass ratios of chemical reactions, scientists can determine how much water is needed to sustain a certain amount of biomass or predict how much oxygen a given plant population releases into the atmosphere. These calculations serve as a fundamental tool for managing ecosystems and designing efficient synthetic processes.