Answer
Approximately 19.3 grams of water will be produced from 12.0 liters of oxygen.
Solution
To solve this problem, we need to follow these steps:
1. **Balance the chemical reaction.**
2. **Determine the amount of water produced from the given volume of oxygen.**
### Step 1: Balance the Chemical Reaction
The unbalanced reaction is:
\[
\mathrm{C}_{3}\mathrm{H}_{3} + \mathrm{O}_{2} \rightarrow \mathrm{H}_{2}\mathrm{O}
\]
To balance this reaction, we need to determine the coefficients for each compound.
1. **Count the atoms of each element:**
- On the left side, we have 3 carbon (C) and 3 hydrogen (H) from \(\mathrm{C}_{3}\mathrm{H}_{3}\) and some oxygen (O) from \(\mathrm{O}_{2}\).
- On the right side, we will have water (\(\mathrm{H}_{2}\mathrm{O}\)), which contains 2 hydrogen and 1 oxygen.
2. **Balance the carbon atoms:**
- We need 3 carbon atoms on the right side, so we place a coefficient of 3 in front of \(\mathrm{H}_{2}\mathrm{O}\):
\[
\mathrm{C}_{3}\mathrm{H}_{3} + \mathrm{O}_{2} \rightarrow 3\mathrm{H}_{2}\mathrm{O}
\]
3. **Balance the hydrogen atoms:**
- Now we have 6 hydrogen atoms on the right side (from \(3\mathrm{H}_{2}\mathrm{O}\)), so we need to balance that by placing a coefficient of 3 in front of \(\mathrm{C}_{3}\mathrm{H}_{3}\):
\[
3\mathrm{C}_{3}\mathrm{H}_{3} + \mathrm{O}_{2} \rightarrow 9\mathrm{H}_{2}\mathrm{O}
\]
4. **Balance the oxygen atoms:**
- On the right side, we have 9 oxygen atoms from \(9\mathrm{H}_{2}\mathrm{O}\). Since \(\mathrm{O}_{2}\) provides 2 oxygen atoms per molecule, we need 4.5 \(\mathrm{O}_{2}\) molecules:
\[
3\mathrm{C}_{3}\mathrm{H}_{3} + 4.5\mathrm{O}_{2} \rightarrow 9\mathrm{H}_{2}\mathrm{O}
\]
5. **Convert to whole numbers:**
- To avoid fractions, we can multiply the entire equation by 2:
\[
6\mathrm{C}_{3}\mathrm{H}_{3} + 9\mathrm{O}_{2} \rightarrow 18\mathrm{H}_{2}\mathrm{O}
\]
The balanced equation is:
\[
6\mathrm{C}_{3}\mathrm{H}_{3} + 9\mathrm{O}_{2} \rightarrow 18\mathrm{H}_{2}\mathrm{O}
\]
### Step 2: Calculate the Amount of Water Produced
Now, we need to find out how many grams of water will be produced from 12.0 liters of oxygen.
1. **Use the ideal gas law to find moles of \(\mathrm{O}_{2}\):**
- At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 liters.
\[
\text{Moles of } \mathrm{O}_{2} = \frac{12.0 \text{ L}}{22.4 \text{ L/mol}} \approx 0.536 \text{ moles}
\]
2. **Use the stoichiometry of the balanced equation:**
- From the balanced equation, 9 moles of \(\mathrm{O}_{2}\) produce 18 moles of \(\mathrm{H}_{2}\mathrm{O}\). Therefore, the ratio is:
\[
\frac{18 \text{ moles } \mathrm{H}_{2}\mathrm{O}}{9 \text{ moles } \mathrm{O}_{2}} = 2 \text{ moles } \mathrm{H}_{2}\mathrm{O} \text{ per mole } \mathrm{O}_{2}
\]
- Thus, the moles of water produced from 0.536 moles of \(\mathrm{O}_{2}\) is:
\[
\text{Moles of } \mathrm{H}_{2}\mathrm{O} = 0.536 \text{ moles } \mathrm{O}_{2} \times 2 = 1.072 \text{ moles } \mathrm{H}_{2}\mathrm{O}
\]
3. **Convert moles of water to grams:**
- The molar mass of water (\(\mathrm{H}_{2}\mathrm{O}\)) is approximately 18 g/mol.
\[
\text{Mass of } \mathrm{H}_{2}\mathrm{O} = 1.072 \text{ moles} \times 18 \text{ g/mol} \approx 19.296 \text{ grams}
\]
### Final Answer
Approximately 19.3 grams of water will be produced from 12.0 liters of oxygen.
Answered by UpStudy AI and reviewed by a Professional Tutor
Explain
Simplify this solution
