Final answer:
To calculate the mass of 1,1,2,2-tetrabromo propane from one gram of bromine, you need to apply stoichiometric principles and the yield percentage but more details are needed to provide an exact answer.
Step-by-step explanation:
The question is concerning the yield of a chemical reaction, specifically the bromination of propene to form 1,1,2,2-tetrabromo propane. When one gram of bromine is used in the reaction at a 27% yield, the mass of 1,1,2,2-tetrabromo propane obtained can be calculated by using stoichiometry and considering the molecular weights of the reactants and products.
To calculate the amount of 1,1,2,2-tetrabromo propane from one gram of bromine, we first need to find the molar mass of bromine (Br2), which is approximately 160 g/mol. Knowing the yield percentage (27%) and the stoichiometry of the reaction (which dictates that one mole of propene reacts with four moles of bromine to form one mole of the product), we can calculate the theoretical yield and then apply the percentage to get the actual yield.
However, the information provided is not sufficient to determine the exact mass of 1,1,2,2-tetrabromo propane produced since the molar mass of 1,1,2,2-tetrabromo propane is not given and neither is the stoichiometry of the reaction with propene. Additional information is required for the calculation.