Discover new knowledge and insights with IDNLearn.com's extensive Q&A platform. Get accurate and timely answers to your queries from our extensive network of experienced professionals.
Sagot :
To determine how many grams of oxygen are required to produce 37.15 grams of carbon dioxide, we can follow these steps:
### Step 1: Determine the number of moles of CO[tex]\(_2\)[/tex]
First, we need to calculate the number of moles of CO[tex]\(_2\)[/tex] produced. The molar mass of CO[tex]\(_2\)[/tex] is 44.01 grams per mole.
[tex]\[ \text{Moles of } CO_2 = \frac{\text{Mass of } CO_2}{\text{Molar mass of } CO_2} = \frac{37.15 \text{ g}}{44.01 \text{ g/mol}} \approx 0.844 \][/tex]
### Step 2: Use the stoichiometric relationship from the balanced equation
From the balanced chemical equation:
[tex]\[ C_3H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O \][/tex]
According to the equation, 1 mole of C[tex]\(_3\)[/tex]H[tex]\(_8\)[/tex] reacts with 5 moles of O[tex]\(_2\)[/tex] to produce 3 moles of CO[tex]\(_2\)[/tex]. Thus, for every 3 moles of CO[tex]\(_2\)[/tex] formed, 5 moles of O[tex]\(_2\)[/tex] are required. We can use this ratio to find the moles of O[tex]\(_2\)[/tex] needed for the given moles of CO[tex]\(_2\)[/tex]:
[tex]\[ \text{Moles of } O_2 = \left(\frac{5 \text{ moles } O_2}{3 \text{ moles } CO_2}\right) \times 0.844 \text{ moles } CO_2 \approx 1.407 \text{ moles } O_2 \][/tex]
### Step 3: Convert the moles of O[tex]\(_2\)[/tex] to grams
Next, we need to convert the moles of O[tex]\(_2\)[/tex] to grams. The molar mass of O[tex]\(_2\)[/tex] is 32.00 grams per mole.
[tex]\[ \text{Mass of } O_2 = \text{Moles of } O_2 \times \text{Molar mass of } O_2 = 1.407 \text{ moles } \times 32.00 \text{ g/mol} \approx 45.02 \text{ g} \][/tex]
### Conclusion
Therefore, 45.02 grams of oxygen are required to produce 37.15 grams of carbon dioxide.
### Step 1: Determine the number of moles of CO[tex]\(_2\)[/tex]
First, we need to calculate the number of moles of CO[tex]\(_2\)[/tex] produced. The molar mass of CO[tex]\(_2\)[/tex] is 44.01 grams per mole.
[tex]\[ \text{Moles of } CO_2 = \frac{\text{Mass of } CO_2}{\text{Molar mass of } CO_2} = \frac{37.15 \text{ g}}{44.01 \text{ g/mol}} \approx 0.844 \][/tex]
### Step 2: Use the stoichiometric relationship from the balanced equation
From the balanced chemical equation:
[tex]\[ C_3H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O \][/tex]
According to the equation, 1 mole of C[tex]\(_3\)[/tex]H[tex]\(_8\)[/tex] reacts with 5 moles of O[tex]\(_2\)[/tex] to produce 3 moles of CO[tex]\(_2\)[/tex]. Thus, for every 3 moles of CO[tex]\(_2\)[/tex] formed, 5 moles of O[tex]\(_2\)[/tex] are required. We can use this ratio to find the moles of O[tex]\(_2\)[/tex] needed for the given moles of CO[tex]\(_2\)[/tex]:
[tex]\[ \text{Moles of } O_2 = \left(\frac{5 \text{ moles } O_2}{3 \text{ moles } CO_2}\right) \times 0.844 \text{ moles } CO_2 \approx 1.407 \text{ moles } O_2 \][/tex]
### Step 3: Convert the moles of O[tex]\(_2\)[/tex] to grams
Next, we need to convert the moles of O[tex]\(_2\)[/tex] to grams. The molar mass of O[tex]\(_2\)[/tex] is 32.00 grams per mole.
[tex]\[ \text{Mass of } O_2 = \text{Moles of } O_2 \times \text{Molar mass of } O_2 = 1.407 \text{ moles } \times 32.00 \text{ g/mol} \approx 45.02 \text{ g} \][/tex]
### Conclusion
Therefore, 45.02 grams of oxygen are required to produce 37.15 grams of carbon dioxide.
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.