When hydrogen is removed from an alkane, an alkene is formed - Edexcel - GCSE Chemistry - Question 10 - 2022 - Paper 1

To maximize the yield and rate of propene production from propane, the manufacturer should consider the following conditions:

1. High Temperature: Increasing the temperature will favor the endothermic reaction, shifting the equilibrium to the right, which increases the formation of propene.

2. Low Pressure: Since there are more moles of gas on the right side of the equilibrium equation (one mole of propane to two moles of products), reducing the pressure will favor the side with more moles, thus increasing the yield of propene.

3. Removing Propene: Continuously removing propene from the reaction mixture will shift the equilibrium to the right, according to Le Chatelier's Principle, thus driving the reaction forward.

4. Using a Catalyst: A catalyst could be employed to increase the rate of reaction without affecting the position of the equilibrium, enabling the manufacturer to produce propene more quickly.

According to the stoichiometry of the reaction, 1 mole of propane produces 1 mole of propene. Therefore, if 300 dm³ of propane is dehydrogenated, the maximum volume of propene formed will also be 300 dm³.

From the reaction, dehydrogenation of 900 dm³ of propane yields 900 dm³ of hydrogen. Using the molar volume of gas at room temperature and pressure (24 dm³/mol), we find the moles of hydrogen:

$ext{Moles of } H_2 = \frac{900 \text{ dm}^3}{24 \text{ dm}^3/\text{mol}} = 37.5 \text{ mol}$

The mass of hydrogen is given by: $ext{Mass} = ext{Moles} \times \text{Molar Mass} = 37.5 \text{ mol} \times 1.0 \text{ g/mol} = 37.5 \text{ g}$

To convert grams to kilograms: $ext{Mass in kg} = \frac{37.5 \text{ g}}{1000} = 0.0375 \text{ kg}$

Thus, the maximum mass of hydrogen formed is 0.0375 kg.