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What is the standard enthalpy change for the reaction of calcium oxide (solid) and water to form calcium hydroxide (solid) if 3.2 grams of calcium oxide reacts with excess water under standard conditions?

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To calculate the standard enthalpy change for the reaction, we need to use the following equation:H = q / nwhere H is the standard enthalpy change, q is the heat released or absorbed during the reaction, and n is the number of moles of the limiting reactant.First, we need to write the balanced chemical equation for the reaction:CaO  s  + H2O  l   Ca OH 2  s Now, we need to find the number of moles of calcium oxide  CaO  in 3.2 grams:Molar mass of CaO = 40.08 g/mol  Ca  + 16.00 g/mol  O  = 56.08 g/moln CaO  = mass / molar mass = 3.2 g / 56.08 g/mol = 0.0571 molNext, we need to find the standard enthalpy change for the reaction. The standard enthalpy of formation for calcium oxide  CaO  is -635.1 kJ/mol, and for calcium hydroxide  Ca OH 2  is -985.2 kJ/mol. Using Hess's Law, we can calculate the standard enthalpy change for the reaction:H reaction  = Hf products  - Hf reactants H reaction  = [1 mol   -985.2 kJ/mol ] - [1 mol   -635.1 kJ/mol ]H reaction  = -985.2 kJ/mol + 635.1 kJ/mol = -350.1 kJ/molNow, we can calculate the heat released  q  during the reaction:q = n  H reaction  = 0.0571 mol   -350.1 kJ/mol  = -20.0 kJSince the value is negative, it means the reaction is exothermic and releases heat.Finally, we can calculate the standard enthalpy change for the reaction:H = q / n = -20.0 kJ / 0.0571 mol = -350.1 kJ/molThe standard enthalpy change for the reaction of calcium oxide and water to form calcium hydroxide is -350.1 kJ/mol.

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