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What is the enthalpy change of precipitation when 50 mL of 0.1 M BaCl2 is mixed with 25 mL of 0.2 M Na2SO4, given that ΔH˚f of BaSO4 is -1464 kJ/mol?

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To calculate the enthalpy change of precipitation, we first need to determine the limiting reactant and the number of moles of the product  BaSO4  formed.The balanced chemical equation for the reaction is:BaCl2 aq  + Na2SO4 aq   BaSO4 s  + 2NaCl aq First, we'll find the moles of each reactant:Moles of BaCl2 =  Volume  Molarity  =  50 mL  0.1 M  = 5 mmolMoles of Na2SO4 =  Volume  Molarity  =  25 mL  0.2 M  = 5 mmolNow, we'll determine the limiting reactant by comparing the mole ratio:Mole ratio of BaCl2 to Na2SO4 = 1:1Since both reactants have the same number of moles and the mole ratio is 1:1, neither reactant is in excess, and both are limiting reactants. Therefore, 5 mmol of BaSO4 will be formed.Now, we can calculate the enthalpy change of precipitation:H = moles of BaSO4  Hf of BaSO4H = 5 mmol   -1464 kJ/mol Since we have the moles in mmol, we need to convert kJ/mol to kJ/mmol:H = 5 mmol   -1.464 kJ/mmol  = -7.32 kJThe enthalpy change of precipitation when 50 mL of 0.1 M BaCl2 is mixed with 25 mL of 0.2 M Na2SO4 is -7.32 kJ.

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