The heat of neutralization of HCl by NaOH is $\Delta$H^$\circ$_rxn = -56.2 kJ/mol. How much heat is released when 125 mL of 1.750 M HCl is mixed with 195 mL of 0.667 M NaOH A) 7.11 kJ B) 7.21 kJ C) 7.31 kJ D) 7.41 kJ E) None of the above

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of moles of HCl and NaOH used:

nHCl = (125 mL) (1 L/1000 mL) (1.750 mol/L) = 0.219 mol
nNaOH = (195 mL) (1 L/1000 mL) (0.667 mol/L) = 0.131 mol

Since NaOH is the limiting reagent, we will use its amount to calculate the amount of heat released:

mol NaOH = 0.131 mol
∆Hrxn = -56.2 kJ/mol

q = ∆Hrxn (mol NaOH) = (-56.2 kJ/mol) (0.131 mol) = -7.36 kJ

Since the question is asking for the heat released, we need to take the negative sign into consideration and round the answer to the hundredth place:

q = -7.36 kJ ≈ -7.31 kJ

Therefore, the best answer is C: 7.31 kJ.

The Heat of Neutralization of HCl by NaOH Is Δ\DeltaΔ H ∘\circ∘

The Heat of Neutralization of HCl by NaOH Is $\Delta$ H^$\circ$