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Answer :
To solve this problem, we need to determine how many grams of water vapor can be produced with the given amounts of each reactant. We can do this by following these steps:
1. Write the balanced chemical equation:
[tex]\[
C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O
\][/tex]
2. Identify the molar masses:
- Molar mass of propane ([tex]\(C_3H_8\)[/tex]): 44.1 g/mol
- Molar mass of oxygen ([tex]\(O_2\)[/tex]): 32.0 g/mol
- Molar mass of water ([tex]\(H_2O\)[/tex]): 18.0 g/mol
3. Calculate the moles of each reactant:
- Moles of [tex]\(C_3H_8\)[/tex]:
[tex]\[
\frac{37.1 \text{ grams}}{44.1 \text{ g/mol}} \approx 0.841 \text{ moles}
\][/tex]
- Moles of [tex]\(O_2\)[/tex]:
[tex]\[
\frac{37.1 \text{ grams}}{32.0 \text{ g/mol}} \approx 1.159 \text{ moles}
\][/tex]
4. Determine the theoretical yield of water from each reactant:
- Propane ([tex]\(C_3H_8\)[/tex]): According to the balanced equation, 1 mole of [tex]\(C_3H_8\)[/tex] will produce 4 moles of [tex]\(H_2O\)[/tex].
[tex]\[
0.841 \text{ moles } C_3H_8 \times 4 \text{ moles } H_2O/\text{mole } C_3H_8 = 3.365 \text{ moles } H_2O
\][/tex]
- Oxygen ([tex]\(O_2\)[/tex]): According to the balanced equation, 5 moles of [tex]\(O_2\)[/tex] produce 4 moles of [tex]\(H_2O\)[/tex].
[tex]\[
\frac{1.159 \text{ moles } O_2}{5} \times 4 = 0.928 \text{ moles } H_2O
\][/tex]
5. Identify the limiting reactant:
- The limiting reactant will produce the least amount of [tex]\(H_2O\)[/tex], therefore [tex]\(O_2\)[/tex] is the limiting reactant because it produces only 0.928 moles of [tex]\(H_2O\)[/tex] compared to 3.365 moles from [tex]\(C_3H_8\)[/tex].
6. Calculate the mass of water produced from the limiting reactant:
[tex]\[
0.928 \text{ moles } H_2O \times 18.0 \text{ g/mol} = 16.695 \text{ grams}
\][/tex]
Thus, the maximum amount of water vapor that can be produced is approximately 16.695 grams.
1. Write the balanced chemical equation:
[tex]\[
C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O
\][/tex]
2. Identify the molar masses:
- Molar mass of propane ([tex]\(C_3H_8\)[/tex]): 44.1 g/mol
- Molar mass of oxygen ([tex]\(O_2\)[/tex]): 32.0 g/mol
- Molar mass of water ([tex]\(H_2O\)[/tex]): 18.0 g/mol
3. Calculate the moles of each reactant:
- Moles of [tex]\(C_3H_8\)[/tex]:
[tex]\[
\frac{37.1 \text{ grams}}{44.1 \text{ g/mol}} \approx 0.841 \text{ moles}
\][/tex]
- Moles of [tex]\(O_2\)[/tex]:
[tex]\[
\frac{37.1 \text{ grams}}{32.0 \text{ g/mol}} \approx 1.159 \text{ moles}
\][/tex]
4. Determine the theoretical yield of water from each reactant:
- Propane ([tex]\(C_3H_8\)[/tex]): According to the balanced equation, 1 mole of [tex]\(C_3H_8\)[/tex] will produce 4 moles of [tex]\(H_2O\)[/tex].
[tex]\[
0.841 \text{ moles } C_3H_8 \times 4 \text{ moles } H_2O/\text{mole } C_3H_8 = 3.365 \text{ moles } H_2O
\][/tex]
- Oxygen ([tex]\(O_2\)[/tex]): According to the balanced equation, 5 moles of [tex]\(O_2\)[/tex] produce 4 moles of [tex]\(H_2O\)[/tex].
[tex]\[
\frac{1.159 \text{ moles } O_2}{5} \times 4 = 0.928 \text{ moles } H_2O
\][/tex]
5. Identify the limiting reactant:
- The limiting reactant will produce the least amount of [tex]\(H_2O\)[/tex], therefore [tex]\(O_2\)[/tex] is the limiting reactant because it produces only 0.928 moles of [tex]\(H_2O\)[/tex] compared to 3.365 moles from [tex]\(C_3H_8\)[/tex].
6. Calculate the mass of water produced from the limiting reactant:
[tex]\[
0.928 \text{ moles } H_2O \times 18.0 \text{ g/mol} = 16.695 \text{ grams}
\][/tex]
Thus, the maximum amount of water vapor that can be produced is approximately 16.695 grams.
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