Certainly! Let's solve the given problem step by step.
### Step 1: Write Down the Balanced Chemical Equation
The balanced chemical equation for photosynthesis is:
[tex]\[ 6 \; \text{CO}_2 + 6 \; \text{H}_2\text{O} \rightarrow \; \text{C}_6\text{H}_{12}\text{O}_6 + 6 \; \text{O}_2 \][/tex]
This equation tells us that 6 moles of water (H[tex]\(_2\)[/tex]O) are needed to produce 1 mole of glucose (C[tex]\(_6\)[/tex]H[tex]\(_{12}\)[/tex]O[tex]\(_6\)[/tex]).
### Step 2: Determine the Moles of Glucose Produced
According to the given information, we start with 3.00 moles of water.
Using the mole ratio from the balanced equation:
- 6 moles of H[tex]\(_2\)[/tex]O produce 1 mole of C[tex]\(_6\)[/tex]H[tex]\(_{12}\)[/tex]O[tex]\(_6\)[/tex].
Given:
[tex]\[ \text{moles of H}_2\text{O} = 3.00 \][/tex]
We can calculate the moles of glucose:
[tex]\[ \text{moles of glucose} = \frac{\text{moles of H}_2\text{O}}{6} \][/tex]
[tex]\[ \text{moles of glucose} = \frac{3.00}{6} \][/tex]
[tex]\[ \text{moles of glucose} = 0.5 \][/tex]
### Step 3: Calculate the Mass of Glucose Produced
To find the mass of glucose produced, we need to know the molar mass of glucose (C[tex]\(_6\)[/tex]H[tex]\(_{12}\)[/tex]O[tex]\(_6\)[/tex]). The molar mass of glucose is given as 180 g/mol.
Given:
[tex]\[ \text{molar mass of glucose} = 180 \; \text{g/mol} \][/tex]
Now, we can calculate the mass using the molar mass and the moles of glucose:
[tex]\[ \text{mass of glucose} = \text{moles of glucose} \times \text{molar mass of glucose} \][/tex]
[tex]\[ \text{mass of glucose} = 0.5 \times 180 \; \text{g/mol} \][/tex]
[tex]\[ \text{mass of glucose} = 90 \; \text{grams} \][/tex]
### Conclusion
Therefore, the mass of glucose produced when 3.00 moles of water react with carbon dioxide is:
[tex]\[ 90 \; \text{grams} \][/tex]
So, the correct answer is:
[tex]\[ \boxed{90 \; \text{grams}} \][/tex]
