a. Biodiesel is an alternative to standard diesel fuel - VCE - SSCE Chemistry - Question 7 - 2010 - Paper 1

First, we calculate the number of moles of POP in 10.0 kg:

$n(\text{POP}) = \frac{10.0 \text{ kg} \times 1000 \text{ g/kg}}{833 \text{ g/mol}} \approx 12.0 \text{ mol}$

Since the stoichiometric ratio is 3:1 (methanol to POP), the number of moles of methanol needed is:

$n(\text{methanol}) = 3 \times n(\text{POP}) = 3 \times 12.0 \text{ mol} = 36.0 \text{ mol}$

Next, we find the volume of methanol required using its density:

$\text{Volume} = \frac{\text{Mass}}{\text{Density}}$

First, calculate the mass of methanol:

$\text{Mass of methanol} = n(\text{methanol}) \times M(\text{methanol})$

The molar mass of methanol (CH₃OH) is approximately 32.04 g/mol, thus:

$\text{Mass of methanol} = 36.0 \text{ mol} \times 32.04 \text{ g/mol} \approx 1150.4 \text{ g}$

Now convert grams to liters:

$\text{Volume} = \frac{1150.4 \text{ g}}{0.79 \text{ g/mL}} \approx 1458.2 \text{ mL} \approx 1.458 \text{ L}$\n Therefore, approximately 1.458 liters of methanol is required.

First, calculate the number of moles of cervonic acid reacted:

$n(\text{cervonic acid}) = \frac{0.328 \text{ g}}{328.0 \text{ g/mol}} = 0.00100 \text{ mol}$

Given that 20.00 mL of 0.300 M I₂ solution was used:

$n(I₂) = 0.300 \text{ mol/L} \times 0.02000 \text{ L} = 0.00600 \text{ mol}$

The reaction indicates that each double bond reacts with one mole of I₂. Therefore, the number of double bonds in cervonic acid is:

$\text{Number of } C=C \text{ double bonds} = \frac{0.00600 \text{ mol I₂}}{0.00100 \text{ mol cervonic acid}} = 6$\n Thus, there are 6 double bonds in a molecule of cervonic acid.

Based on the information provided, the molecular formula for cervonic acid can be inferred knowing that it contains 22 carbon atoms and 6 double bonds. The general formula can be stated as:

$C_{22}H_{34}O_{2}$

as it accounts for the double bonds while also satisfying the hydrogen balance with respect to the carbon count.