The diagram shows the structure of a propene molecule - Edexcel - GCSE Chemistry - Question 3 - 2011 - Paper 1

To illustrate the polymer formation, draw two propene molecules aligned such that their double bonds interact, breaking to form a single bond. Then, represent the repeating unit of the polymer, indicating that the structure continues similarly, showcasing the loss of a small molecule (like H$_2$) for each bond formed.

One common test for the presence of carbon-carbon double bonds is the addition reaction with bromine water. When propene is mixed with bromine water, the brown color of bromine disappears, indicating that the double bond has reacted and formed a dibrominated product.

To determine the total mass of the products, we first need to establish the molar mass of decane (C$_{10}$H$_{22}$), which is approximately 142 g/mol. The stoichiometry shows that 1 mole of decane produces 2 moles of propene and 1 mole of butane. Hence, from 17 g of decane, we can find that:

1. Calculate moles of decane: Moles = mass / molar mass = 17 g / 142 g/mol ≈ 0.119 moles.
2. The total moles of products formed = 2(0.119) + 0.119 = 0.357 moles.
3. Calculate the mass using the total molar mass of the products (propene and butane) to find the mass of the total products formed.

Cracking refers to the process of breaking down larger hydrocarbon molecules into smaller, more useful ones, typically through the application of heat and sometimes in the presence of a catalyst. This process allows the conversion of less valuable fractions into more desirable products such as alkenes and smaller alkanes.

Cracking is necessary because:

1. There is generally less demand for fractions containing large molecules which are less useful as fuels or raw materials.
2. Large molecules are inefficient as fuels; cracking produces smaller, more widely used alkanes and alkenes which serve various applications, such as in fuels and to create plastics.