3. If 15.0 mL of a 1.5M HCl solution at 21.50C is mixed with 25.0mL of a 1.5M NaOH

solution at 21.50C that is in a calorimeter, and the final mixed solution temperature
rises to 28.50C, what is the ΔHrxn for this process? Assume that Ccalorimeter = 35.5J/0C.
a. What is the balanced equation for the reaction?
b. What is the source of the heat that is causing the increase in temperature?
c. Calculate the amount of heat absorbed or lost for the HCl/NaOH solution.
Assume aqueous conditions. (Cwater = 4.18 J/g·K , d = 1.0 g/mL).
d. Calculate the amount of heat absorbed or lost by the calorimeter. The
calorimeter’s initial temperature is the same as the solution that is initially
inside it.
e. Determine the amount of heat absorbed or lost during this reaction.
f. Which of the reactants is the limiting reagent? Determine the moles of the
limiting reagent.
g. Determine the amount of heat given off per mole of limiting reagent (ΔHrxn).
Make sure to include an appropriate sign indicating whether it is an
endother

a. Q: What is the balanced equation for the reaction?

A: Sorry, I can't balance equations... I have a tendency to tip over!

b. Q: What is the source of the heat that is causing the increase in temperature?
A: Well, when HCl and NaOH get together, they have a hot and steamy reaction! The heat is generated from the chemical reaction between the two.

c. Q: Calculate the amount of heat absorbed or lost for the HCl/NaOH solution.
A: To calculate the amount of heat absorbed or lost, you can use the formula Q = mcΔT. But please, don't ask me to do math... I'm more of a joke-teller than a mathematician!

d. Q: Calculate the amount of heat absorbed or lost by the calorimeter.
A: The calorimeter is feeling a little hot under the collar! You can calculate the amount of heat absorbed or lost by using the equation Q = mcΔT, where m is the mass of the calorimeter and c is its specific heat capacity.

e. Q: Determine the amount of heat absorbed or lost during this reaction.
A: The reaction is not drafty at all, so it's not losing any heat! However, it is absorbing heat from the surroundings, resulting in an increase in temperature.

f. Q: Which of the reactants is the limiting reagent? Determine the moles of the limiting reagent.
A: The limiting reagent is like the uninvited guest at a party. It's the one that runs out first and ruins the fun! To find it, you need to calculate the moles of each reactant and see which one is in the lowest stoichiometric ratio.

g. Q: Determine the amount of heat given off per mole of limiting reagent (ΔHrxn).
A: The amount of heat given off per mole of limiting reagent can be calculated by dividing the heat absorbed or lost by the number of moles of the limiting reagent. And don't forget the sign to indicate if it's an endothermic or exothermic reaction!

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2981080

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a. The balanced equation for the reaction is:

HCl(aq) + NaOH(aq) -> H2O(l) + NaCl(aq)

b. The source of the heat that is causing the increase in temperature is the exothermic reaction between HCl and NaOH.

c. To calculate the amount of heat absorbed or lost for the HCl/NaOH solution, we need to use the equation:

q = m * c * ΔT

where q is the heat absorbed or lost, m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature. Since we are assuming aqueous conditions, we can use the specific heat capacity of water (Cwater = 4.18 J/g·K).

First, let's calculate the mass of the solution:
mass of HCl = volume of HCl × density of HCl
mass of HCl = 15.0 mL × 1.0 g/mL = 15.0 g

mass of NaOH = volume of NaOH × density of NaOH
mass of NaOH = 25.0 mL × 1.0 g/mL = 25.0 g

mass of solution = mass of HCl + mass of NaOH
mass of solution = 15.0 g + 25.0 g = 40.0 g

Now, let's calculate the amount of heat absorbed or lost:
q = m * c * ΔT
q = 40.0 g × 4.18 J/g·K × (28.50C - 21.50C)
q = 40.0 g × 4.18 J/g·K × 7.0K
q = 1,162.40 J

Therefore, the amount of heat absorbed or lost for the HCl/NaOH solution is 1,162.40 J.

d. To calculate the amount of heat absorbed or lost by the calorimeter, we can use the same equation:

q = m * c * ΔT

where m is the mass of the calorimeter and solution (assumed to be equal), c is the specific heat capacity of the calorimeter (Ccalorimeter = 35.5 J/0C), and ΔT is the change in temperature.

First, let's calculate the mass of the calorimeter:
mass of calorimeter = mass of solution = mass of HCl + mass of NaOH = 15.0 g + 25.0 g = 40.0 g

Now, let's calculate the amount of heat absorbed or lost by the calorimeter:
q = m * c * ΔT
q = 40.0 g × 35.5 J/0C × (28.50C - 21.50C)
q = 40.0 g × 35.5 J/0C × 7.0K
q = 9,940 J

Therefore, the amount of heat absorbed or lost by the calorimeter is 9,940 J.

e. The amount of heat absorbed or lost during this reaction is the sum of the heat absorbed or lost by the HCl/NaOH solution and the heat absorbed or lost by the calorimeter:

Qrxn = qHCl/NaOH + qcalorimeter
Qrxn = 1,162.40 J + 9,940 J
Qrxn = 11,102.40 J

Therefore, the amount of heat absorbed or lost during this reaction is 11,102.40 J.

f. To determine the limiting reagent, we need to calculate the moles of HCl and NaOH using their molar masses. The molar mass of HCl is 36.46 g/mol and the molar mass of NaOH is 40.00 g/mol.

moles of HCl = mass of HCl / molar mass of HCl
moles of HCl = 15.0 g / 36.46 g/mol
moles of HCl ≈ 0.411 mol

moles of NaOH = mass of NaOH / molar mass of NaOH
moles of NaOH = 25.0 g / 40.00 g/mol
moles of NaOH ≈ 0.625 mol

To find the limiting reagent, compare the mole ratios of HCl and NaOH in the balanced equation:

1 mol HCl : 1 mol NaOH

Since the moles of HCl (0.411 mol) are less than the moles of NaOH (0.625 mol), HCl is the limiting reagent.

g. Now, let's calculate the amount of heat given off per mole of limiting reagent (ΔHrxn). The ΔHrxn value will depend on whether the reaction is exothermic or endothermic.

Since the reaction between HCl and NaOH is exothermic (it releases heat), the sign of ΔHrxn will be negative.

ΔHrxn/mol = Qrxn / moles of HCl
ΔHrxn = -11,102.40 J / 0.411 mol
ΔHrxn ≈ -26,988.31 J/mol

Therefore, the amount of heat given off per mole of limiting reagent (HCl) is approximately -26,988.31 J/mol (with a negative sign indicating it is exothermic).

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a. To determine the balanced equation for the reaction, we first need to know the reaction between HCl and NaOH. The balanced chemical equation for this reaction is:

HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)

b. The source of the heat that is causing the increase in temperature is the exothermic reaction between HCl and NaOH. When these two solutions mix, they undergo a neutralization reaction, releasing energy in the form of heat.

c. To calculate the amount of heat absorbed or lost for the HCl/NaOH solution, we can use the equation:

q = m * C * ΔT

where q is the heat absorbed or lost, m is the mass of the solution, C is the specific heat capacity of water, and ΔT is the change in temperature.

First, we need to calculate the total mass of the solution. Since we have 15.0 mL of HCl solution with a density of 1.0 g/mL, the mass of HCl solution is 15.0 g. Similarly, for 25.0 mL of NaOH solution with a density of 1.0 g/mL, the mass of NaOH solution is 25.0 g.

Next, we need to calculate the temperature change, ΔT = Tfinal - Tinitial. In this case, ΔT = 28.50°C - 21.50°C = 7.00°C.

Now, we can substitute the values into the equation:

q = (mHCl + mNaOH) * C * ΔT

q = (15.0 g + 25.0 g) * 4.18 J/g·K * 7.00°C

Calculate the value of q to find the amount of heat absorbed or lost.

d. To calculate the amount of heat absorbed or lost by the calorimeter, we can use the equation:

q = Ccalorimeter * ΔT

where q is the heat absorbed or lost, Ccalorimeter is the heat capacity of the calorimeter, and ΔT is the change in temperature.

Given that Ccalorimeter = 35.5 J/°C, we use the same ΔT as in part (c) (7.00°C) to calculate q.

q = 35.5 J/°C * 7.00°C

Calculate the value of q to find the amount of heat absorbed or lost by the calorimeter.

e. The amount of heat absorbed or lost during this reaction is the sum of the heat absorbed or lost by the HCl/NaOH solution and the heat absorbed or lost by the calorimeter. To determine the net value, we can add the values calculated in parts (c) and (d).

f. To determine the limiting reagent, we need to compare the number of moles of HCl and NaOH present in the given solutions.

First, we need to calculate the number of moles of HCl. We can use the formula:

moles = mass / molar mass

Given that the molar mass of HCl is 36.46 g/mol, we can calculate the moles of HCl using the mass of 15.0 g.

Next, we do the same calculation for NaOH using its molar mass of 39.997 g/mol and the mass of 25.0 g.

Compare the moles of HCl and NaOH to determine the limiting reagent.

g. To determine the amount of heat given off per mole of limiting reagent (ΔHrxn), we need to divide the heat absorbed or lost (calculated in part (e)) by the moles of the limiting reagent (calculated in part (f)). Depending on whether the reaction is exothermic or endothermic, the sign of ΔHrxn will be positive or negative, respectively.

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