Question

Determine the standard enthalpy change for the precipitation reaction of 10.0 g of calcium chloride (CaCl2) with excess sodium carbonate (Na2CO3) to form calcium carbonate (CaCO3) and sodium chloride (NaCl). Given the following standard enthalpies of formation: ΔHf°(CaCl2) = -795.8 kJ/mol, ΔHf°(Na2CO3) = -1130.1 kJ/mol, ΔHf°(CaCO3) = -1206.9 kJ/mol, and ΔHf°(NaCl) = -411.2 kJ/mol.

Answer 1

First, we need to write the balanced chemical equation for the reaction:CaCl2  aq  + Na2CO3  aq   CaCO3  s  + 2 NaCl  aq Next, we need to determine the moles of CaCl2 in 10.0 g:Molar mass of CaCl2 = 40.08  Ca  + 2 * 35.45  Cl  = 110.98 g/molMoles of CaCl2 = 10.0 g / 110.98 g/mol = 0.0901 molNow, we can use the standard enthalpies of formation to calculate the standard enthalpy change for the reaction:H reaction  = [Hf CaCO3  + 2 * Hf NaCl ] - [Hf CaCl2  + Hf Na2CO3 ]H reaction  = [ -1206.9 kJ/mol  + 2 *  -411.2 kJ/mol ] - [ -795.8 kJ/mol  +  -1130.1 kJ/mol ]H reaction  =  -1206.9 - 822.4 + 795.8 + 1130.1  kJ/molH reaction  = -103.4 kJ/molFinally, we need to multiply the standard enthalpy change for the reaction by the moles of CaCl2 to find the total enthalpy change for the 10.0 g of CaCl2:Total enthalpy change = H reaction  * moles of CaCl2Total enthalpy change = -103.4 kJ/mol * 0.0901 mol = -9.32 kJTherefore, the standard enthalpy change for the precipitation reaction of 10.0 g of calcium chloride with excess sodium carbonate is -9.32 kJ.