What is the enthalpy change when 49.4 mL of 0.430 M sulfuric acid reacts with 23.3 mL of 0.309 M potassium hydroxide? H₂SO₄(aq) + 2KOH(aq) $\to$ K₂SO₄(aq) + 2H₂O(l)
$\Delta$ H° = -111.6 kJ/mol

Question

What is the enthalpy change when 49.4 mL of 0.430 M sulfuric acid reacts with 23.3 mL of 0.309 M potassium hydroxide? H₂SO₄(aq)+ 2KOH(aq)$\to$ K₂SO₄(aq)+ 2H₂O(l) $\Delta$H° = -111.6 kJ/mol

Quizplus · Accepted Answer

First, we need to find the limiting reactant. 
Number of moles of sulfuric acid: 
0.430 mol/L x 0.0494 L = 0.021257 mol 
Number of moles of potassium hydroxide: 
0.309 mol/L x 0.0233 L = 0.007204 mol 
Since we need 2 moles of potassium hydroxide for every 1 mole of sulfuric acid, the limiting reactant is potassium hydroxide. 
Number of moles of sulfuric acid used: 
0.007204 mol x 1 mol sulfuric acid/2 mol potassium hydroxide = 0.003602 mol 
Enthalpy change: 
-111.6 kJ/mol x 0.003602 mol = -0.4019 kJ 
Since the enthalpy change is negative, it means the reaction is exothermic. The answer rounded to three significant figures is -0.402 kJ. Therefore, the best answer choice is A.

What Is the Enthalpy Change When 49 →\to→ K2SO4(aq)+ 2H2O(l) Δ\DeltaΔ H° = -1116 KJ/mol A)-0

What Is the Enthalpy Change When 49 $\to$ K₂SO₄(aq)+ 2H₂O(l) $\Delta$ H° = -1116 KJ/mol
A)-0