- The balanced chemical equation \(6 CO_2 + 6 H_2O \rightarrow C_6H_{12}O_6 + 6 O_2\) shows that 6 moles of \(O_2\) are produced for every 1 mole of \(C_6H_{12}O_6\).
- To convert moles of \(O_2\) to moles of \(C_6H_{12}O_6\), we need a mole ratio that places \(O_2\) in the denominator to cancel existing units.
- The correct ratio is \(\frac{1 \text{ mol } C_6H_{12}O_6}{6 \text{ mol } O_2}\), which corresponds to option (a).
- The final answer is \(\boxed{\text{a}}\).
Explanation
- Analyzing the Chemical Reaction and Stoichiometry
The balanced chemical equation for photosynthesis is \(6 CO_2 + 6 H_2O \rightarrow C_6H_{12}O_6 + 6 O_2\). This equation tells us the stoichiometric relationships between the reactants and the products. Specifically, it states that 6 moles of carbon dioxide (\(CO_2\)) and 6 moles of water (\(H_2O\)) react to produce 1 mole of glucose (\(C_6H_{12}O_6\)) and 6 moles of oxygen (\(O_2\)). To solve the given stoichiometry problem, we focus on the mole ratio between oxygen (\(O_2\)) and glucose (\(C_6H_{12}O_6\)), which is 6 moles of \(O_2\) for every 1 mole of \(C_6H_{12}O_6\).
- Determining the Required Mole Ratio
We are performing a dimensional analysis calculation to convert the mass of \(O_2\) to the mass of \(C_6H_{12}O_6\). The expression is given as:
\[1.00 \text{ g } O_2 \times \frac{1 \text{ mol } O_2}{32.00 \text{ g } O_2} \times \frac{?}{?} \times \frac{180.18 \text{ g } C_6H_{12}O_6}{1 \text{ mol } C_6H_{12}O_6} = x \text{ g } C_6H_{12}O_6\]
To correctly convert from moles of \(O_2\) to moles of \(C_6H_{12}O_6\), we need a mole ratio where moles of \(O_2\) are in the denominator (to cancel the units from the previous term) and moles of \(C_6H_{12}O_6\) are in the numerator. Based on the balanced chemical equation, the ratio is \(\frac{1 \text{ mol } C_6H_{12}O_6}{6 \text{ mol } O_2}\).
- Selecting the Correct Ratio
By examining the provided options:
A\(\frac{1 \text{ mol } C_6H_{12}O_6}{6 \text{ mol } O_2}\)
B\(\frac{6 \text{ mol } C_6H_{12}O_6}{1 \text{ mol } O_2}\)
C\(\frac{1 \text{ mol } O_2}{6 \text{ mol } C_6H_{12}O_6}\)
D\(\frac{6 \text{ mol } O_2}{1 \text{ mol } C_6H_{12}O_6}\)
We see that option (a) matches our derived ratio of \(\frac{1 \text{ mol } C_6H_{12}O_6}{6 \text{ mol } O_2}\). This correctly allows for the cancellation of 'mol \(O_2\)' and conversion to 'mol \(C_6H_{12}O_6\)' in our dimensional analysis chain. Therefore, the correct mole ratio is (a). The final answer is \(\boxed{\text{a}}\).
Examples
Stoichiometry is fundamental in industrial chemistry and laboratory experiments. For instance, if you are a chemical engineer managing the production of a specific substance, you must use precise mole ratios derived from balanced equations to determine how much of each raw material is needed. By mastering these ratios, you ensure that processes are efficient, waste is minimized, and products are synthesized correctly, which is crucial for everything from manufacturing medicine to generating clean energy.