The molar mass of glycerol, C3H8O3, is 92.0 g mol–1 - VCE - SSCE Chemistry - Question 10 - 2022 - Paper 1

To determine the amount of water produced during the synthesis of 65.0 g of C3H6O3 from the triglyceride C51H98O6, we need to analyze the reaction.

The reaction involves the hydrolysis of the triglyceride, resulting in the release of glycerol (C3H8O3) and fatty acids (from tripalmitin).

The general equation for each triglyceride can be described as:

$C_3H_8O_3 + 3 ext{ Fatty Acids} \rightarrow C_3H_6O_3 + 3 H_2O$

To find the moles of water produced, we first need to calculate the moles of glycerol produced:

1. Calculate moles of glycerol:

   • Molar mass of glycerol = 92.0 g/mol
   • Moles of glycerol produced = ( \frac{65.0 \text{ g}}{92.0 \text{ g/mol}} \approx 0.707 \text{ mol} )

2. From the balanced equation, 1 mol of glycerol produces 3 mol of water:

   • Moles of water produced = ( 0.707 \text{ mol} \times 3 \approx 2.121 \text{ mol} )

3. Convert moles of water to grams:

   • Molar mass of water = 18.0 g/mol
   • Grams of water = ( 2.121 \text{ mol} \times 18.0 \text{ g/mol} \approx 38.2 ext{ g} )

Thus, the reaction requires 38.2 g of water, which corresponds to option B.