Final answer:
The mass of 2-bromopropane prepared from 25.5 g of propene, assuming a 100% yield, is greater than 25.5 g, because the molecular weight of 2-bromopropane is higher due to the bromine atom.
Step-by-step explanation:
The question asks us to calculate the mass of 2-bromopropane that could be prepared from 25.5 g of propene, assuming a 100% yield of the product. To solve this, we need to use the molecular weights of propene (C3H6) and 2-bromopropane (C3H7Br).
First, we calculate the number of moles of propene using its molecular weight (42.08 g/mol):
25.5 g propene x (1 mol propene / 42.08 g propene) = 0.606 moles propene
Since the reaction assumes a 100% yield, the moles of propene will equal the moles of 2-bromopropane produced. Now, we use the molecular weight of 2-bromopropane (122.96 g/mol) to find the mass:
0.606 moles 2-bromopropane x (122.96 g 2-bromopropane / 1 mol 2-bromopropane) = 74.48 g 2-bromopropane
Therefore, less than 25.5 g (option c) is incorrect because the correct mass of 2-bromopropane formed would be higher due to the addition of a bromine atom.