Answer :
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.
