Answer :
To solve this problem, we need to determine how many grams of water vapor will be produced when 9.21 grams of carbon dioxide are produced during cellular respiration. Let’s go through the solution step-by-step.
### Step 1: Understand the molecular equation
The balanced equation for cellular respiration is:
[tex]\[ C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O \][/tex]
This tells us that one mole of glucose (C₆H₁₂O₆) reacts with six moles of oxygen (O₂) to produce six moles of carbon dioxide (CO₂) and six moles of water (H₂O).
### Step 2: Calculate moles of CO₂ produced
First, we need to know how many moles of CO₂ are produced from 9.21 grams of CO₂.
- The molar mass of CO₂ (carbon dioxide) is calculated as follows:
[tex]\[ \text{Molar mass of } CO_2 = 12.01 \frac{\text{g}}{\text{mol}} (\text{C}) + 2 \times 16.00 \frac{\text{g}}{\text{mol}} (\text{O}) = 44.01 \frac{\text{g}}{\text{mol}} \][/tex]
Given mass of CO₂ is 9.21 grams.
[tex]\[ \text{Moles of CO}_2 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{9.21 \, \text{g}}{44.01 \, \frac{\text{g}}{\text{mol}}} \approx 0.20927062031356514 \, \text{mol} \][/tex]
### Step 3: Calculate moles of H₂O produced
According to the balanced equation, the molar ratio of CO₂ to H₂O is 1:1. Therefore, the moles of H₂O produced will be the same as the moles of CO₂ produced.
[tex]\[ \text{Moles of H}_2\text{O} = \text{Moles of CO}_2 = 0.20927062031356514 \, \text{mol} \][/tex]
### Step 4: Calculate the mass of H₂O produced
To find the mass of water produced, we need to convert moles of water to grams.
- The molar mass of H₂O (water) is calculated as follows:
[tex]\[ \text{Molar mass of H}_2O = 2 \times 1.01 \frac{\text{g}}{\text{mol}} (\text{H}) + 16.00 \frac{\text{g}}{\text{mol}} (\text{O}) = 18.015 \frac{\text{g}}{\text{mol}} \][/tex]
[tex]\[ \text{Mass of H}_2\text{O} = \text{Moles of H}_2\text{O} \times \text{Molar mass of H}_2\text{O} = 0.20927062031356514 \, \text{mol} \times 18.015 \, \frac{\text{g}}{\text{mol}} \approx 3.770010224948876 \, \text{g} \][/tex]
### Conclusion
Therefore, if an organism produces 9.21 grams of carbon dioxide from cellular respiration, approximately 3.77 grams of water vapor will also be formed.
### Step 1: Understand the molecular equation
The balanced equation for cellular respiration is:
[tex]\[ C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O \][/tex]
This tells us that one mole of glucose (C₆H₁₂O₆) reacts with six moles of oxygen (O₂) to produce six moles of carbon dioxide (CO₂) and six moles of water (H₂O).
### Step 2: Calculate moles of CO₂ produced
First, we need to know how many moles of CO₂ are produced from 9.21 grams of CO₂.
- The molar mass of CO₂ (carbon dioxide) is calculated as follows:
[tex]\[ \text{Molar mass of } CO_2 = 12.01 \frac{\text{g}}{\text{mol}} (\text{C}) + 2 \times 16.00 \frac{\text{g}}{\text{mol}} (\text{O}) = 44.01 \frac{\text{g}}{\text{mol}} \][/tex]
Given mass of CO₂ is 9.21 grams.
[tex]\[ \text{Moles of CO}_2 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{9.21 \, \text{g}}{44.01 \, \frac{\text{g}}{\text{mol}}} \approx 0.20927062031356514 \, \text{mol} \][/tex]
### Step 3: Calculate moles of H₂O produced
According to the balanced equation, the molar ratio of CO₂ to H₂O is 1:1. Therefore, the moles of H₂O produced will be the same as the moles of CO₂ produced.
[tex]\[ \text{Moles of H}_2\text{O} = \text{Moles of CO}_2 = 0.20927062031356514 \, \text{mol} \][/tex]
### Step 4: Calculate the mass of H₂O produced
To find the mass of water produced, we need to convert moles of water to grams.
- The molar mass of H₂O (water) is calculated as follows:
[tex]\[ \text{Molar mass of H}_2O = 2 \times 1.01 \frac{\text{g}}{\text{mol}} (\text{H}) + 16.00 \frac{\text{g}}{\text{mol}} (\text{O}) = 18.015 \frac{\text{g}}{\text{mol}} \][/tex]
[tex]\[ \text{Mass of H}_2\text{O} = \text{Moles of H}_2\text{O} \times \text{Molar mass of H}_2\text{O} = 0.20927062031356514 \, \text{mol} \times 18.015 \, \frac{\text{g}}{\text{mol}} \approx 3.770010224948876 \, \text{g} \][/tex]
### Conclusion
Therefore, if an organism produces 9.21 grams of carbon dioxide from cellular respiration, approximately 3.77 grams of water vapor will also be formed.