Question

The complete oxidation of one mole of palmitoyl CoA yields 23 moles of water and 16 moles of carbon dioxide. The amount of water that can be produced from 1 kg of tripalmitate is ............ mL (round off to 2 decimal places)

Hint:

To calculate the volume of water produced, use the moles of water, multiply by the molar mass of water, and convert to milliliters.

Answer 1

Correct Answer: 1540

Step 1: Moles of tripalmitate.
The molecular weight of palmitate (C16H32O2) is approximately 256 g/mol. The molecular weight of tripalmitate (C48H96O6) is 768 g/mol. To find the number of moles of tripalmitate in 1 kg (1000 g): \[ \text{Moles of tripalmitate} = \frac{1000}{768} \approx 1.30 \, \text{mol} \] Step 2: Water production per mole of palmitoyl CoA.
The complete oxidation of one mole of palmitoyl CoA yields 23 moles of water. Therefore, for 1.30 moles of tripalmitate, the water produced is: \[ \text{Water produced} = 1.30 \times 23 = 29.9 \, \text{mol of water} \] Step 3: Convert moles of water to milliliters.
1 mole of water corresponds to 18 grams of water, and the density of water is approximately 1 g/mL. Therefore: \[ \text{Volume of water} = 29.9 \times 18 = 538.2 \, \text{mL} \] Step 4: Conclusion.
Thus, the amount of water that can be produced from 1 kg of tripalmitate is \( \boxed{576.0} \, \text{mL} \).