A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO₃ in a coffee cup calorimeter.If both solutions were initially at 35.00°C and the temperature of the resulting solution was recorded as 37.00°C,determine the ΔH°_rxn (in units of kJ/mol NaOH)for the neutralization reaction between aqueous NaOH and HCl.Assume 1)that no heat is lost to the calorimeter or the surroundings,and 2)that the density and the heat capacity of the resulting solution are the same as water. A) -55.7 kJ/mol NaOH B) -169 kJ/mol NaOH C) -16.7 kJ/mol NaOH D) -27.9 kJ/mol NaOH E) - 34.4 kJ/mol NaOH

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Quizplus · Accepted Answer

The heat of neutralization can be calculated using the equation:ΔH°rxn = - (q / n)where q is the heat released or absorbed and n is the number of moles of the limiting reactant.In this case, the limiting reactant is NaOH, and the number of moles of NaOH is:n = (0.300 mol/L) * (0.100 L) = 0.0300 molThe heat released is:q = (37.00°C - 35.00°C) * (100.0 mL + 100.0 mL) * (1.00 g/mL) * (4.184 J/g°C) = 836.8 JTherefore, the ΔH°rxn is:ΔH°rxn = - (836.8 J / 0.0300 mol) = -27.9 kJ/mol NaOH

A 1000 ML Sample of 0