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2 years ago

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When 13.6 g of calcium chloride, CaCl2, was dissolved in 100.0 mL of water in a coffee cup calorimeter, the temperature rose fro

m 25.00 °C to 31.75 °C. Determine the ∆Hrxn in kJ/mol for the reaction as written below. Assume that the density of the solutions is 1.000 g/mL, and that the specific heat capacity of the solutions is that of pure water. Must show your work (scratch paper) to receive credit.

1 answer:

DanielleElmas [232]2 years ago

5 0

Answer:

THE ENTHALPY OF SOLUTION IS 3153.43 J/MOL OR 3.15 KJ/MOL.

Explanation:

1. write out the variables given:

Mass of Calcium chloride = 13.6 g

Change in temperature = 31.75°C - 25.00°C = 6.75 °C

Density of the solution = 1.000 g/mL

Volume = 100.0 mL = 100.0 mL

Specific heat of water = 4.184 J/g °C

Mass of the water = unknown

2. calculate the mass of waterinvolved:

We must first calculate the mass of water in the bomb calorimeter

Mass = density * volume

Mass = 1.000 * 100

Mass = 0.01 g

3. calculate the quantity of heat evolved:

Next is to calculate the quantity of heat evolved from the reaction

Heat = mass * specific heat of water * change in temperature

Heat = mass of water * specific heat *change in temperature

Heat = 13.6 g * 4.184 * 6.75

Heat = 13.6 g * 4.184 J/g °C * 6.75 °C

Heat = 384.09 J

Hence, 384.09J is the quantity of heat involved in the reaction of 13.6 g of calcium chloride in the calorimeter.

4. calculate the molar mass of CaCl2:

Next is to calculate the molar mas of CaCl2

Molar mass = ( 40 + 35.5 *2) = 111 g/mol

The number of moles of 13.6 g of CaCl2 is then:

Number of moles of CaCl2 = mass / molar mass

Number of moles = 13.6 g / 111 g/mol

Number of moles = 0.1225 mol

So 384.09 J of heat was involved in the reaction of 1.6 g of CaCl2 in a calorimter which translates to 0.1225 mol of CaCl2..

5. Calculate the enthalpy of solution in kJ/mol:

If 1 mole of CaCl2 is involved, the heat evolved is therefore:

Heat per mole = 384.09 J / 0.1225 mol

Heat = 3 135.43 J/mol

The enthalpy of solution is therefore 3153.43 J/mol or 3.15 kJ/mol.

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2 years ago

vichka [17]

Answer:

1. 0.1683 mol

2. 1.191 g

3. 0.02695 mol

4. Na₂Cl₃

5. The empirical formula obtained is not correct. This is likely due to experimental errors, since much precision was required (the masses are determined in thousandths of grams).

Explanation:

<em><u></u></em>

<em><u>1. How many moles of elemental sodium were used in the reaction?</u></em>

Since all of the solid sodium is used up by the reaction, you can cancluate the number of moles of elemental sodium used dividing the mass by the molar mass:

number of moles = mass in grams / atomic mass

mass in grams = 0.3870 g (given)

atomic mass = 22.990 g/mol

number of moles = 0.3870 g / 22.990 g/mol = 0.1683 mol

<u><em>2. What is the mass of chlorine gas used in the reaction?</em></u>

a) Mass of chlorine gas introduced in the flask = mass of the stoppered flask after filling it with chlorine gas - mass of the empty flask with the sopper

Mass of chlorine gas introduced = 158.1743g - 156.1870g = 1.9873 g

b) Mass of chlorine gas unreacted = 0.7962 g (given)

c) Mass of chlorine gas used = mass of chorine gas introduced in the flask - mass of chlorine gass un reacted

Mass of chlorine gas used = 1.9873g - 0.7962g = 1.1911g

<u><em>3. How many moles of chlorine were used in the reaction?</em></u>

molar mass of chlorine gas, Cl₂ = 2 × 35.453 g/mol = 70.906 g/mol

number of moles = mass in grams / molar mass = 1.911g / 70.906g/mol = 0.02695 mol

<u><em>4. What is the empirical formula of sodium chloride based on the experimental data?</em></u>

Divide the number of moles of each element by the smalles number of moles:

Na: 0.01683 / 0.01683 = 1

Cl = 0.02695 / 0.01683 = 1.6

Multiply by 2 to obtain whole numbers:

Na = 2

Cl = 3.2 = 3

Empirical formula Na₂Cl₃

<u><em></em></u>

<u><em>5. Was the empirical formula you obtained correct using the chemists data correct? Why? </em></u>

<u><em></em></u>

No, the empirical formula you obtained using the chemists data is not correct, because the correct empirical formula of sodium chloride is NaCl.

That is, there is 1 atom of sodium per every atom of chlorine in one chemical formula of NaCl, but that is not reflected by the empirical formula Na₂Cl₃.

That is a demostration of big experimental errors. You can speculate that the errors are likely due to problems of procedure collecting the gas or errors in measuring the masses.

As you see, the masses are measured to thousandths of grams, which requires much precision; thus smalls absolute errors could produce huge relative errors.

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The chemical equation for the combustion of hydrocarbon having carbon and hydrogen follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

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<u>For calculating the mass of carbon:</u>

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So, in 0.3986 g of carbon dioxide, $\frac{12}{44}\times 0.3986=0.1087g$ of carbon will be contained.

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In 18g of water, 2 g of hydrogen is contained.

So, in 0.0578 g of water, $\frac{2}{18}\times 0.0578=0.0066g$ of hydrogen will be contained.

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Mass of sample = 0.1153 g

Mass of carbon = 0.1087 g

Putting values in equation 1, we get:

$\%\text{ composition of carbon}=\frac{0.1087g}{0.1153g}\times 100=94.27\%$

<u>For Hydrogen:</u>

Mass of sample = 0.1153 g

Mass of hydrogen = 0.0066 g

Putting values in equation 1, we get:

$\%\text{ composition of hydrogen}=\frac{0.0066g}{0.1153g}\times 100=5.72\%$

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