Question

Consider the combination reaction between aluminum metal and liquid bromine (d = 3.1028 g/mL). Also consider that you have 38.4 g of aluminum metal and 40.6 g of bromine liquid. For this reaction AHrxn = -526 kJ/mol. If you were able to measure 15.5 kJ of heat being dissipated from this reaction, what would you expect to be the percent yield based on the heat generated? Hint: Find the limiting reagent before calculating the theoretical yield of the reaction.

Answer 1

The balanced equation is 2 Al 3 Br2 2 AIBr3 -------- Number of moles of Al 38.4 g/(26.98 g/mol) = 1.4233 mol Number of moles of Br2 40.6 g/(159.81 g/mol) = 0.2541 mol According to balanced equation, 3 moles of Br2 reacts with 2 moles of Al So, 0.2541 mol of Br2 would react with 0.2541 mol * 2/3 0.169368 mol of Al But, the number of moles of Al present is more than required. Hence Br2 is the Limiting reagent. Accordingto balanced equation, 3 moles of Br2 produces 2 moles of AlBr3 So, 0.2541 mol of Br2 would produce 0.2541 mol 2/3 0.169368 mol of AlBr3 Number of moles of AlBr3 produced 0.169368 mol Enthalpy of reaction - 526 kJ/mol 0.169368 mol - 89.09 kJ Actual heat produced 15.5 kJ Percent yield 15.5/89.09 * 100 = 17.40 %