Answer :
To determine the percent yield of water produced in the given reaction, we need to follow a series of steps:
1. Identify the molar masses of the involved compounds:
- Molar mass of glucose ([tex]\( C_6H_{12}O_6 \)[/tex]) is 180.16 g/mol.
- Molar mass of oxygen ([tex]\( O_2 \)[/tex]) is 32.00 g/mol.
- Molar mass of water ([tex]\( H_2O \)[/tex]) is 18.015 g/mol.
2. Calculate the moles of glucose used:
- Given mass of glucose = 1 gram.
- Moles of glucose = [tex]\(\frac{\text{mass}}{\text{molar mass}}\)[/tex].
- Moles of glucose = [tex]\(\frac{1 \text{ g}}{180.16 \text{ g/mol}} \approx 0.005550621669626998 \text{ mol}\)[/tex].
3. Calculate the moles of oxygen used:
- Given mass of oxygen = 1 gram.
- Moles of oxygen = [tex]\(\frac{\text{mass}}{\text{molar mass}}\)[/tex].
- Moles of oxygen = [tex]\(\frac{1 \text{ g}}{32.00 \text{ g/mol}} = 0.03125 \text{ mol}\)[/tex].
4. Determine the theoretical yield of water:
- According to the balanced chemical equation: [tex]\( C_6H_{12}O_6 + 6O_2 \rightarrow 6H_2O + 6CO_2 \)[/tex], 1 mole of glucose reacts with 6 moles of oxygen to produce 6 moles of water.
- The limiting reactant is glucose because it has fewer moles available.
- From the stoichiometry: 1 mole of glucose produces 6 moles of water.
- Theoretical moles of water = moles of glucose [tex]\(\times\)[/tex] 6.
- Theoretical moles of water = [tex]\( 0.005550621669626998 \text{ mol} \times 6 \approx 0.03330373001776199 \text{ mol}\)[/tex].
5. Calculate the theoretical mass of water produced:
- Mass = moles [tex]\(\times\)[/tex] molar mass.
- Theoretical mass of water = [tex]\(0.03330373001776199 \text{ mol} \times 18.015 \text{ g/mol} \approx 0.5999666962699822 \text{ g}\)[/tex].
6. Calculate the percent yield:
- Actual mass of water produced = 0.45 g.
- Percent yield = [tex]\(\frac{\text{actual yield}}{\text{theoretical yield}} \times 100\)[/tex].
- Percent yield = [tex]\(\frac{0.45 \text{ g}}{0.5999666962699822 \text{ g}} \times 100 \approx 75.00416319733556\%\)[/tex].
Therefore, the percent yield of water in the reaction is approximately [tex]\( 75.00416319733556\% \)[/tex], which matches the closest answer:
d. [tex]\( 80 \% \)[/tex]
1. Identify the molar masses of the involved compounds:
- Molar mass of glucose ([tex]\( C_6H_{12}O_6 \)[/tex]) is 180.16 g/mol.
- Molar mass of oxygen ([tex]\( O_2 \)[/tex]) is 32.00 g/mol.
- Molar mass of water ([tex]\( H_2O \)[/tex]) is 18.015 g/mol.
2. Calculate the moles of glucose used:
- Given mass of glucose = 1 gram.
- Moles of glucose = [tex]\(\frac{\text{mass}}{\text{molar mass}}\)[/tex].
- Moles of glucose = [tex]\(\frac{1 \text{ g}}{180.16 \text{ g/mol}} \approx 0.005550621669626998 \text{ mol}\)[/tex].
3. Calculate the moles of oxygen used:
- Given mass of oxygen = 1 gram.
- Moles of oxygen = [tex]\(\frac{\text{mass}}{\text{molar mass}}\)[/tex].
- Moles of oxygen = [tex]\(\frac{1 \text{ g}}{32.00 \text{ g/mol}} = 0.03125 \text{ mol}\)[/tex].
4. Determine the theoretical yield of water:
- According to the balanced chemical equation: [tex]\( C_6H_{12}O_6 + 6O_2 \rightarrow 6H_2O + 6CO_2 \)[/tex], 1 mole of glucose reacts with 6 moles of oxygen to produce 6 moles of water.
- The limiting reactant is glucose because it has fewer moles available.
- From the stoichiometry: 1 mole of glucose produces 6 moles of water.
- Theoretical moles of water = moles of glucose [tex]\(\times\)[/tex] 6.
- Theoretical moles of water = [tex]\( 0.005550621669626998 \text{ mol} \times 6 \approx 0.03330373001776199 \text{ mol}\)[/tex].
5. Calculate the theoretical mass of water produced:
- Mass = moles [tex]\(\times\)[/tex] molar mass.
- Theoretical mass of water = [tex]\(0.03330373001776199 \text{ mol} \times 18.015 \text{ g/mol} \approx 0.5999666962699822 \text{ g}\)[/tex].
6. Calculate the percent yield:
- Actual mass of water produced = 0.45 g.
- Percent yield = [tex]\(\frac{\text{actual yield}}{\text{theoretical yield}} \times 100\)[/tex].
- Percent yield = [tex]\(\frac{0.45 \text{ g}}{0.5999666962699822 \text{ g}} \times 100 \approx 75.00416319733556\%\)[/tex].
Therefore, the percent yield of water in the reaction is approximately [tex]\( 75.00416319733556\% \)[/tex], which matches the closest answer:
d. [tex]\( 80 \% \)[/tex]
