Answer :
To solve this problem, we need to determine how many grams of carbon dioxide ([tex]\(CO_2\)[/tex]) are produced when 30 liters of oxygen ([tex]\(O_2\)[/tex]) are reacted at standard temperature and pressure (STP).
### Step-by-step Solution:
1. Understand the Chemical Equation:
The balanced chemical equation given is:
[tex]\[
C_5H_{12} + 8O_2 \rightarrow 5CO_2 + 6H_2O
\][/tex]
From this equation, we can see that 8 moles of [tex]\(O_2\)[/tex] react to produce 5 moles of [tex]\(CO_2\)[/tex].
2. Determine the Molar Volume at STP:
At STP (standard temperature and pressure), the volume occupied by one mole of any gas is 22.4 liters.
3. Calculate Moles of Oxygen ([tex]\(O_2\)[/tex]):
We are given 30 liters of [tex]\(O_2\)[/tex]. To find moles of [tex]\(O_2\)[/tex], use the molar volume:
[tex]\[
\text{moles of } O_2 = \frac{\text{volume of } O_2}{\text{molar volume at STP}} = \frac{30 \, \text{liters}}{22.4 \, \text{liters/mol}} \approx 1.339 \, \text{moles}
\][/tex]
4. Calculate Moles of Carbon Dioxide ([tex]\(CO_2\)[/tex]):
From the balanced equation, 8 moles of [tex]\(O_2\)[/tex] produce 5 moles of [tex]\(CO_2\)[/tex]. Therefore, the moles of [tex]\(CO_2\)[/tex] produced are:
[tex]\[
\text{moles of } CO_2 = \left(\frac{5}{8}\right) \times \text{moles of } O_2 = \left(\frac{5}{8}\right) \times 1.339 \approx 0.837 \, \text{moles}
\][/tex]
5. Calculate Grams of Carbon Dioxide ([tex]\(CO_2\)[/tex]):
The molar mass of [tex]\(CO_2\)[/tex] is 44 g/mol. Therefore, the mass of [tex]\(CO_2\)[/tex] produced is:
[tex]\[
\text{grams of } CO_2 = \text{moles of } CO_2 \times \text{molar mass of } CO_2 = 0.837 \, \text{moles} \times 44 \, \text{g/mol} \approx 36.83 \, \text{grams}
\][/tex]
Thus, when 30 liters of [tex]\(O_2\)[/tex] are reacted, approximately 36.83 grams of [tex]\(CO_2\)[/tex] will be produced. This rounded value doesn't match exactly any of the given options, but the closest rounding would be choice a) 94, although the computed answer isn't directly listed. Please ensure you've listed all possible answer choices correctly.
### Step-by-step Solution:
1. Understand the Chemical Equation:
The balanced chemical equation given is:
[tex]\[
C_5H_{12} + 8O_2 \rightarrow 5CO_2 + 6H_2O
\][/tex]
From this equation, we can see that 8 moles of [tex]\(O_2\)[/tex] react to produce 5 moles of [tex]\(CO_2\)[/tex].
2. Determine the Molar Volume at STP:
At STP (standard temperature and pressure), the volume occupied by one mole of any gas is 22.4 liters.
3. Calculate Moles of Oxygen ([tex]\(O_2\)[/tex]):
We are given 30 liters of [tex]\(O_2\)[/tex]. To find moles of [tex]\(O_2\)[/tex], use the molar volume:
[tex]\[
\text{moles of } O_2 = \frac{\text{volume of } O_2}{\text{molar volume at STP}} = \frac{30 \, \text{liters}}{22.4 \, \text{liters/mol}} \approx 1.339 \, \text{moles}
\][/tex]
4. Calculate Moles of Carbon Dioxide ([tex]\(CO_2\)[/tex]):
From the balanced equation, 8 moles of [tex]\(O_2\)[/tex] produce 5 moles of [tex]\(CO_2\)[/tex]. Therefore, the moles of [tex]\(CO_2\)[/tex] produced are:
[tex]\[
\text{moles of } CO_2 = \left(\frac{5}{8}\right) \times \text{moles of } O_2 = \left(\frac{5}{8}\right) \times 1.339 \approx 0.837 \, \text{moles}
\][/tex]
5. Calculate Grams of Carbon Dioxide ([tex]\(CO_2\)[/tex]):
The molar mass of [tex]\(CO_2\)[/tex] is 44 g/mol. Therefore, the mass of [tex]\(CO_2\)[/tex] produced is:
[tex]\[
\text{grams of } CO_2 = \text{moles of } CO_2 \times \text{molar mass of } CO_2 = 0.837 \, \text{moles} \times 44 \, \text{g/mol} \approx 36.83 \, \text{grams}
\][/tex]
Thus, when 30 liters of [tex]\(O_2\)[/tex] are reacted, approximately 36.83 grams of [tex]\(CO_2\)[/tex] will be produced. This rounded value doesn't match exactly any of the given options, but the closest rounding would be choice a) 94, although the computed answer isn't directly listed. Please ensure you've listed all possible answer choices correctly.
