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What is the heat of neutralization of sulfuric acid (H2SO4) with sodium hydroxide (NaOH) at constant pressure if 50 mL of 2.0 M H2SO4 is reacted with 50 mL of 2.0 M NaOH?

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To calculate the heat of neutralization, we first need to determine the balanced chemical equation for the reaction between sulfuric acid  H2SO4  and sodium hydroxide  NaOH . The balanced equation is:H2SO4 + 2NaOH  Na2SO4 + 2H2ONow, we need to determine the number of moles of each reactant. We can do this using the given concentrations and volumes:moles of H2SO4 =  concentration of H2SO4    volume of H2SO4 moles of H2SO4 =  2.0 mol/L    50 mL  1 L/1000 mL moles of H2SO4 = 0.1 molmoles of NaOH =  concentration of NaOH    volume of NaOH moles of NaOH =  2.0 mol/L    50 mL  1 L/1000 mL moles of NaOH = 0.1 molSince the stoichiometry of the reaction is 1:2  1 mol of H2SO4 reacts with 2 mol of NaOH , we can see that both reactants are present in stoichiometric amounts, so neither is in excess.Now, we can calculate the heat of neutralization. The heat of neutralization for a strong acid and a strong base is approximately -57.32 kJ/mol of water formed. In this reaction, 2 moles of water are formed for every mole of H2SO4 reacted.Heat of neutralization =  moles of H2SO4    heat of neutralization per mole of water formed    number of moles of water formed per mole of H2SO4 Heat of neutralization =  0.1 mol    -57.32 kJ/mol    2 mol Heat of neutralization = -11.464 kJSo, the heat of neutralization of sulfuric acid with sodium hydroxide at constant pressure for the given reaction conditions is approximately -11.464 kJ.

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