Okay, so I need to calculate the enthalpy of neutralization when these three solutions are mixed. Hmm, let's see. The question gives three different solutions: HCl, H2SO4, and Ba(OH)2. I remember that the enthalpy of neutralization is the heat released when an acid reacts with a base to form water. But wait, in this case, there are three different acids and one base. So I need to figure out which ones react with each other and how much of each is present.

First, I should write down the balanced chemical equations for each possible reaction. Let's start with HCl and H2SO4. HCl is a strong monoprotic acid, so it donates one H+ per molecule. H2SO4 is a strong diprotic acid, meaning it can donate two H+ ions. But when mixing with a base like Ba(OH)2, which is a strong base, I need to see which acids react with the base.

Ba(OH)2 is a strong base and will react with both HCl and H2SO4. Let me check the balanced equations:

For HCl and Ba(OH)2:
HCl + Ba(OH)2 → BaCl2 + H2O
The balanced equation is 2 HCl + Ba(OH)2 → BaCl2 + 2 H2O. So the mole ratio is 2:1.

For H2SO4 and Ba(OH)2:
H2SO4 + Ba(OH)2 → BaSO4 + 2 H2O
The balanced equation is H2SO4 + Ba(OH)2 → BaSO4 + 2 H2O. So the mole ratio is 1:1.

Now, I need to calculate the moles of each acid and see how much of each reacts with the base. Let's start with the given volumes and concentrations.

Calculating moles:

HCl: 500 mL = 0.5 L, 0.2 M. Moles = Molarity × Volume = 0.2 * 0.5 = 0.1 mol.

H2SO4: 300 mL = 0.3 L, 0.3 M. Moles = 0.3 * 0.3 = 0.09 mol.

Ba(OH)2: 200 mL = 0.2 L, 0.5 M. Moles = 0.5 * 0.2 = 0.1 mol.

Now, Ba(OH)2 has 0.1 mol. Let's see which reactions it can have.

First, let's consider the reaction with HCl. The balanced equation requires 2 moles of HCl per 1 mole of Ba(OH)2. So 0.1 mol Ba(OH)2 would require 0.2 mol HCl. But we only have 0.1 mol HCl. So HCl is the limiting reactant here. Wait, no. Wait, if Ba(OH)2 reacts with HCl first, but let's check which one is the limiting reagent when considering both possible reactions.

Alternatively, maybe the base reacts with both acids. But the problem is that the acid and base might neutralize each other in a certain order. I need to determine which acid reacts first with the base, or if both react simultaneously. Hmm. But in reality, when you mix all three, the base will react with the available acids. But since HCl and H2SO4 are both acids, the base will react with them in the order of their stoichiometric availability.

Wait, but in a neutralization reaction, the strong acid (HCl) and strong diprotic acid (H2SO4) will both react with the strong base (Ba(OH)2). So the base will react with the acids until it's all consumed. So we need to see how much of each acid reacts with the base.

But the problem is that the base's moles (0.1 mol) can react with both HCl and H2SO4. Let's figure out how much of each acid is consumed by the base.

First, let's see how much HCl can react with Ba(OH)2. The reaction is 2 HCl + Ba(OH)2 → products. So 1 mole of Ba(OH)2 reacts with 2 moles of HCl. We have 0.1 mol Ba(OH)2. So it can react with 0.2 mol HCl. But we only have 0.1 mol HCl. So all the HCl will be consumed by the base. Then, the remaining Ba(OH)2 will react with H2SO4.

Wait, but let's check:

Total Ba(OH)2: 0.1 mol.

If all HCl (0.1 mol) is consumed, then the Ba(OH)2 used in that reaction is 0.05 mol (since 2 HCl : 1 Ba(OH)2 → 0.1 HCl : 0.05 Ba(OH)2). So remaining Ba(OH)2 is 0.1 - 0.05 = 0.05 mol. This remaining Ba(OH)2 will then react with H2SO4.

Each mole of Ba(OH)2 reacts with 1 mole of H2SO4. So 0.05 mol Ba(OH)2 will react with 0.05 mol H2SO4. But we have 0.09 mol H2SO4. So only 0.05 mol of H2SO4 is consumed, and 0.04 mol remains.

So in total, the reaction consumes 0.1 mol HCl and 0.05 mol H2SO4, and 0.1 mol Ba(OH)2. The remaining H2SO4 is 0.04 mol.

But wait, the problem is asking for the enthalpy of neutralization. Enthalpy of neutralization is typically for the reaction between a strong acid and a strong base, producing water. But here, we have two different acids and one base. So the neutralization would involve two separate reactions: one between HCl and Ba(OH)2, and another between the remaining H2SO4 and Ba(OH)2. However, the total heat released would be the sum of the heat released in both reactions.

But wait, the question says "the enthalpy of neutralization when 500 mL 0.2 M HCl, 300 mL 0.3 M H2SO4 and 200 mL 0.5 M Ba(OH)2 are mixed." So it's a single process. So the enthalpy change would be the sum of the heat from both reactions. But wait, actually, the enthalpy of neutralization is usually defined per mole of H+ neutralized. But in this case, since we have two acids reacting with the base, the total moles of H+ neutralized would be the sum from both reactions.

Alternatively, maybe the problem assumes that all the HCl and H2SO4 are neutralized by Ba(OH)2, and we need to calculate the total heat released. Then, the enthalpy of neutralization would be the heat released per mole of H+ neutralized. Wait, but the standard enthalpy of neutralization for strong acid-strong base is -57 kJ/mol (approximately). But when you have different acids, the amount of H+ neutralized varies.

Wait, maybe the problem expects us to calculate the total heat released when all possible neutralization occurs, and then determine the enthalpy per mole of equivalent (moles of H+ neutralized). Let's proceed step by step.

First, calculate the moles of H+ from each acid:

HCl is a strong monoprotic acid, so 0.1 mol HCl → 0.1 mol H+.

H2SO4 is a strong diprotic acid, so 0.09 mol H2SO4 → 0.18 mol H+ (since each H2SO4 donates 2 H+).

Ba(OH)2 is a strong base, 0.5 M * 0.2 L = 0.1 mol Ba(OH)2. Ba(OH)2 dissociates into Ba^2+ and 2 OH^- → 0.2 mol OH^-.

So the total OH^- available is 0.2 mol.

The H+ from the acids are 0.1 (from HCl) + 0.18 (from H2SO4) = 0.28 mol H+.

The OH^- available is 0.2 mol. So all OH^- will react with H+ to form water. The reaction is H+ + OH^- → H2O, ΔH = -57 kJ/mol (approximately for strong acid-strong base neutralization).

Total moles of H+ neutralized: 0.2 mol (since OH^- is 0.2 mol). Wait, but H+ is 0.28 mol. So not all H+ will be neutralized. Only 0.2 mol of H+ will be neutralized by the OH^-.

Therefore, the total heat released would be 0.2 mol * (-57 kJ/mol) = -11.4 kJ. But wait, the problem is asking for the enthalpy of neutralization. Wait, enthalpy of neutralization is typically per mole of equivalent. Each OH^- neutralizes one H+, so the equivalent is 0.2 mol. So the enthalpy change is -11.4 kJ for 0.2 mol of neutralization. But the question might be asking for the enthalpy per mole of reaction (i.e., per mole of H+ neutralized). So per mole of H+ neutralized, it's -57 kJ. But in this case, only 0.2 mol are neutralized. So the total heat released is -11.4 kJ. But the options are in kJ and kcal. Let's check the options.

The options are:

A. -16.0 kJ

B. -3.80 kcal

C. -2.72 kcal

D. -11.42 kcal

Wait, so the total heat released is -11.4 kJ. To convert that to kcal, since 1 kcal ≈ 4.184 kJ, so 11.4 kJ / 4.184 ≈ 2.72 kcal. But since the heat is released, it's exothermic, so the sign is negative. So -11.4 kJ is approximately -2.72 kcal. But looking at the options, option C is -2.72 kcal and option D is -11.42 kcal. Wait, but 11.4 kJ is 11.4 kJ, which is 11.4/4.184 ≈ 2.72 kcal. So the total enthalpy change is -11.4 kJ, which is equivalent to -2.72 kcal. But wait, the question is asking for the enthalpy of neutralization. Wait, but the enthalpy of neutralization is typically per mole of equivalent (i.e., per mole of H+ neutralized). In this case, 0.2 mol of H+ are neutralized, so the total enthalpy change is -11.4 kJ. So the enthalpy per mole of H+ neutralized would be -57 kJ/mol. But the options don't have -57 kJ. Wait, maybe I'm misunderstanding the question.

Wait, the problem says "the enthalpy of neutralization when ... are mixed." So perhaps it's the heat released per mole of neutralization reaction. But since there are multiple acids involved, maybe the total heat is the sum of the neutralizations. Alternatively, perhaps the question expects us to calculate the total heat released, not per mole of equivalent.

Wait, let's re-express the problem. The enthalpy of neutralization is typically defined as the heat released when one mole of H+ from a strong acid reacts with one mole of OH- from a strong base to form water. So in this case, the total moles of H+ neutralized would be the moles of OH- available, which is 0.2 mol. So the total heat released is 0.2 mol * (-57 kJ/mol) = -11.4 kJ. But the options have -11.42 kcal, which is approximately -2.72 kcal. Wait, but -11.4 kJ is approximately -2.72 kcal. But the question's options are in kcal, so maybe the answer is -2.72 kcal, which is option C. However, option D is -11.42 kcal, which is the same as -11.4 kJ converted to kcal. Wait, 11.4 kJ is approximately 2.72 kcal (since 1 kcal = 4.184 kJ). So 11.4 kJ / 4.184 ≈ 2.72 kcal. So -11.4 kJ is -2.72 kcal. So the correct answer would be option C. But looking at the options, option C is -2.72 kcal and option D is -11.42 kcal. Wait, but the problem asks for the correct answer. However, another approach is needed here.

Wait, maybe I made a mistake in calculating the moles of H+ neutralized. Let's re-examine.

Total moles of OH- available from Ba(OH)2: 200 mL * 0.5 M = 0.1 mol Ba(OH)2. Each Ba(OH)2 gives 2 OH- ions, so 0.2 mol OH-.

Total moles of H+ available from the acids: HCl (0.1 mol) + H2SO4 (0.09 mol * 2 = 0.18 mol) = 0.28 mol H+.

OH- is 0.2 mol. So the reaction will consume 0.2 mol H+ and 0.2 mol OH- to form 0.2 mol H2O. The remaining H+ is 0.28 - 0.2 = 0.08 mol. The remaining H2SO4 is 0.09 - 0.05 = 0.04 mol (since each H2SO4 gives 2 H+; but if 0.05 mol Ba(OH)2 reacts with H2SO4, that's 0.05 mol H2SO4 reacted, leaving 0.04 mol). The remaining HCl is 0.1 - 0.1 = 0 mol. The remaining H2SO4 is 0.04 mol, which remains unreacted. However, the unreacted H2SO4 would dissociate into 0.08 mol H+ (since each H2SO4 gives 2 H+). So total remaining H+ is 0.08 mol. But the question is asking for the enthalpy of neutralization, which is the heat released when the neutralization occurs. Since not all H+ is neutralized, only 0.2 mol is. So the heat released is 0.2 mol * (-57 kJ/mol) = -11.4 kJ. Converting to kcal: 11.4 kJ / 4.184 ≈ 2.72 kcal. So the total heat released is -11.4 kJ or -2.72 kcal. But the options are:

A. -16.0 kJ

B. -3.80 kcal

C. -2.72 kcal

D. -11.42 kcal

Wait, but if the question is asking for the enthalpy change per mole of H+ neutralized, then it's -57 kJ/mol. But none of the options have that. Alternatively, if the question is asking for the total enthalpy change for all the neutralization that occurs (i.e., the 0.2 mol), then the answer would be -11.4 kJ or -2.72 kcal. But the options have both -11.42 kcal and -2.72 kcal. But the problem states "the enthalpy of neutralization", which is typically the heat per mole of reaction, which is -57 kJ/mol. However, in this case, not all H+ are neutralized, so the total heat is -11.4 kJ, which is equivalent to -2.72 kcal. So the answer would be option C. But wait, let me check the calculations again.

Wait, maybe I made a mistake in calculating the moles of H+ neutralized. Let's see:

Ba(OH)2 has 0.1 mol. Each mole of Ba(OH)2 provides 2 OH- ions. So total OH- is 0.2 mol.

H+ from HCl: 0.1 mol.

H+ from H2SO4: 0.09 mol * 2 = 0.18 mol.

Total H+ available: 0.28 mol.

OH- available: 0.2 mol.

So OH- will neutralize 0.2 mol H+.

Therefore, the heat released is 0.2 mol * (-57 kJ/mol) = -11.4 kJ.

Converting to kcal: 11.4 kJ / 4.184 ≈ 2.72 kcal.

So the total heat released is -11.4 kJ or -2.72 kcal. But the question is asking for the enthalpy of neutralization. Wait, but enthalpy of neutralization is usually given per mole of H+ neutralized. In this case, 0.2 mol of H+ are neutralized, so the total heat is -11.4 kJ. But the options have -2.72 kcal (option C) and -11.42 kcal (option D). Wait, but -11.4 kJ is approximately -2.72 kcal, so the correct answer would be option C. However, option D is -11.42 kcal, which is the same as -11.4 kJ converted directly (since 1 kJ ≈ 0.239 kcal, so 11.4 kJ * 0.239 ≈ 2.72 kcal). So -11.4 kJ is -2.72 kcal. Therefore, the correct answer is option C. But wait, the options are:

A. -16.0 kJ

B. -3.80 kcal

C. -2.72 kcal

D. -11.42 kcal

Wait, but -11.4 kJ is -11.4 kJ, and -11.42 kcal is approximately -11.42 * 4.184 ≈ -47.2 kJ. That doesn't make sense. Wait, no. Wait, 1 kcal = 4.184 kJ, so to convert kJ to kcal, divide by 4.184. So 11.4 kJ / 4.184 ≈ 2.72 kcal. Therefore, -11.4 kJ is equivalent to -2.72 kcal. So the correct answer is option C. However, option D is -11.42 kcal, which would be the same as -11.42 * 4.184 ≈ -47.2 kJ, which doesn't match. So the correct answer should be option C. But let's check the calculations again.

Alternatively, maybe the problem is considering the neutralization of all available H+ with the available OH-. So total H+ is 0.28 mol, OH- is 0.2 mol. So 0.2 mol H+ are neutralized, 0.08 mol H+ remain. The heat released is 0.2 mol * 57 kJ/mol = 11.4 kJ (exothermic, so negative). Then, converting 11.4 kJ to kcal: 11.4 / 4.184 ≈ 2.72 kcal. So the total heat released is -2.72 kcal, which is option C. But wait, the options have D as -11.42 kcal. That would be if we converted 11.4 kJ directly to kcal (without considering moles of H+). But the enthalpy of neutralization is per mole of H+ neutralized. So the correct answer should be option C. However, looking back at the problem statement, the question says "the enthalpy of neutralization when ... are mixed." The enthalpy of neutralization is usually defined per mole of H+ neutralized, so the answer should be -57 kJ/mol. But in this case, only 0.2 mol were neutralized, so the total heat is -11.4 kJ, which is -2.72 kcal. Therefore, the correct answer is option C. But the options don't have -57 kJ/mol. Alternatively, maybe the problem is asking for the molar enthalpy of neutralization per mole of reaction, considering the total moles of H+ neutralized. Wait, but the standard enthalpy of neutralization is -57 kJ/mol, which is for the reaction of one mole of H+ with one mole of OH-. Here, 0.2 mol are neutralized, so the total heat is 0.2 * -57 = -11.4 kJ, which is -2.72 kcal. Therefore, the correct answer is option C. However, the options have D as -11.42 kcal, which would be if we took 11.4 kJ as kcal (which is incorrect). So the correct answer is C. But wait, let's check the problem again.

The problem says "calculate the enthalpy of neutralization when ... are mixed." The enthalpy of neutralization is the heat released per mole of reaction. If the reaction involves mixing all three solutions, the total heat released would be the sum of the heat from each neutralization step. However, since Ba(OH)2 can only neutralize 0.2 mol H+ (as there's only 0.2 mol OH-), the total heat released is 0.2 * 57 kJ = 11.4 kJ (negative). Converting to kcal: 11.4 / 4.184 ≈ 2.72 kcal. So the answer is -2.72 kcal, which is option C. Therefore, the correct answer is C. However, option D is -11.42 kcal, which would be if we didn't convert units correctly. Alternatively, maybe the problem assumes that all H+ are neutralized, but that's not possible because OH- is only 0.2 mol. So the answer should be C.
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