Question

A student needs to find out how much heat is released or absorbed when 100 mL of 0.1 mol/L solution of barium chloride (BaCl2) is mixed with 150 mL of 0.1 mol/L solution of sodium sulfate (Na2SO4) at 25°C. Given the enthalpy of precipitation of BaSO4 as -256 kJ/mol, calculate the enthalpy change (in kJ) for the reaction that occurs and determine whether the reaction is exothermic or endothermic.

Answer 1

First, we need to determine the balanced chemical equation for the reaction:BaCl2 aq  + Na2SO4 aq   BaSO4 s  + 2NaCl aq Next, we need to find the limiting reactant. To do this, we will calculate the moles of each reactant:Moles of BaCl2 =  volume in L    concentration in mol/L Moles of BaCl2 =  100 mL   1 L/1000 mL    0.1 mol/L = 0.01 molMoles of Na2SO4 =  volume in L    concentration in mol/L Moles of Na2SO4 =  150 mL   1 L/1000 mL    0.1 mol/L = 0.015 molSince there is a 1:1 ratio between BaCl2 and Na2SO4 in the balanced equation, BaCl2 is the limiting reactant because it has fewer moles  0.01 mol  than Na2SO4  0.015 mol .Now, we can calculate the enthalpy change for the reaction using the enthalpy of precipitation of BaSO4:Enthalpy change =  moles of limiting reactant    enthalpy of precipitation per mole Enthalpy change =  0.01 mol    -256 kJ/mol  = -2.56 kJThe enthalpy change for the reaction is -2.56 kJ. Since the enthalpy change is negative, the reaction is exothermic, meaning heat is released during the reaction.