Question

In: Chemistry

1) What volume of 17.4 M acetic acid should be used to produce 0.100 L of...

1) What volume of 17.4 M acetic acid should be used to produce 0.100 L of 3.00 M acetic acid solution?

2) Given the following thermochemical equation, how much total energy is released when 1.50 kg of CH4 is combusted?:
CH4 (g) + 2O2 (g) -----> CO2(g) + 2H2O (l)

Thank you!!

Solutions

Expert Solution

1) Given:

Molarity of acetic acid M1 = 17.4 M

The volume of acetic acid V2 = 0.100 L

Molarity of acetic acid M2 = 3.00 M

Solution:

To find the volume used to produce 0.100 L of 3.00 M acetic acid solution from a stock solution:

The formula M1V1 = M2V2 where “1” represents the concentrated conditions and “2” represents the diluted conditions.

V1 = M2V2/ M1

V1 = (3.00M*0.100L)/17.4 M

V1 = 0.01724 L

V1 = 17.24 mL

17.24 mL of 17.4 M acetic acid should be used to produce 0.100 L of 3.00 M acetic acid solution

------------------------------------------------------------------------------------------------

2) Given the thermochemical equation:

CH4 (g) + 2O2 (g) → CO2(g) + 2H2O (l)

The heat of reaction = [the sum of all heats of formation of all products] - [the sum of all heats of formation of all reactants]

The calculation for the heat of combustion for methane is

The heat of combustion of CH4= [heat of formation of CO2(g) + 2 * heat of formation of H2O (l)] –

            [heat of formation of CH4 (g) + 2 * heat of formation of O2 (g)]

Heat of combustion of CH4= [-393.5 kJ/mol + 2 * (-285.8 kJ/mol)] – [-74.8 kJ/mol + 2 * (0 kJ/mol)]

Heat of combustion of CH4= [-393.5 kJ/mol + (-571.6 kJ/mol)] – [-74.8 kJ/mol + (0 kJ/mol)]

Heat of combustion of CH4= - 965.1 kJ/mol + 74.8 kJ/mol = -890.3 kJ/mol

Negative sign indicates heat is released.

One mole of methane release 890.3 kJ

Calculate the number of moles in 1.50 kg of methane

Molar mass of methane = 16.04 g/mol

Mass of methane = 1.50 kg = 1500 g

Moles of methane = mass of methane / Molar mass of methane

Moles of methane = 1500 g / 16.04 g/mol

Moles of methane = 1500 g / 16.04 g/mol = 93.5162 mol

One mole of methane produces 890.3 kJ of energy

93.5162 moles of methane produce = (890.3 kJ * 93.5162 mol)/1 mol = 83257.4729 kJ

83257.4 kJ of energy released from 1.50 kg of methane is combusted.

Round off three significant figures, then 83300 kJ heat produced.

83300 kJ of energy released from 1.50 kg of methane is combusted.


Related Solutions

To what volume (in mL) should 8.25 mL of an 6.5 M acetic acid solution be...
To what volume (in mL) should 8.25 mL of an 6.5 M acetic acid solution be diluted in order to obtain a final solution that is 2 M?
In the titration of 25.00 mL of 0.100 M acetic acid, what is the pH of...
In the titration of 25.00 mL of 0.100 M acetic acid, what is the pH of the resultant solution after the addition of 15.06 mL of 0.100 M of KOH? Assume that the volumes of the solutions are additive. Ka = 1.8x10-5 for acetic acid, CH3COOH.
In the titration of 25.00 mL of 0.100 M acetic acid, what is the pH of...
In the titration of 25.00 mL of 0.100 M acetic acid, what is the pH of the resultant solution after the addition of 25.00 mL of 0.100 M KOH? Assume that the volumes of the solutions are additive. Ka = 1.8x10-5 for CH3COOH.
Consider the titration of 25.0 mL of 0.100 M acetic acid (HA) with 0.100 M NaOH....
Consider the titration of 25.0 mL of 0.100 M acetic acid (HA) with 0.100 M NaOH. 1. Write the balanced chemical equation and equilibrium constant expression (ECE) for all of the reactions that occur when NaOH is added to the acetic throughout the titration. Hint: think of what is in the solution (acetic acid) with water, acetic acid with sodium hydroxide, and acetate ion with water) as the titration is proceeding. 2. Calculate the volume of NaOH solution needed to...
For the titration of 50.0 mL of 0.150 M acetic acid with 0.100 M sodium hydroxide,...
For the titration of 50.0 mL of 0.150 M acetic acid with 0.100 M sodium hydroxide, determine the pH when: (a) 50.0 mL of base has been added. (b) 75.0 mL of base has been added. (c) 100.0 mL of base has been added.
A 25.0 mL sample of 0.100 M acetic acid is titrated with a 0.125 M NaOH...
A 25.0 mL sample of 0.100 M acetic acid is titrated with a 0.125 M NaOH solution. Calculate the pH of the mixture after 10, 20, and 30 ml of NaOH have been added (Ka=1.76*10^-5)
You are titrating 0.100 L of 0.100 M carbonic acid (shown below) with 0.500 M NaOH....
You are titrating 0.100 L of 0.100 M carbonic acid (shown below) with 0.500 M NaOH. What will the pH be when 23.0 mL of NaOH have been added? pkas: 6.05 and 10.64
A 25.0-mL solution of 0.100 M acetic acid is titrated with a 0.200 M KOH solution....
A 25.0-mL solution of 0.100 M acetic acid is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of the KOH solution: a) 0.0 mL b) 5.0 mL c) 10.0 mL d) 12.5 mL e) 15.0 mL
50.0 ml of an acetic acid (CH3COOH) of unknown concentration is titrated with 0.100 M NaOH....
50.0 ml of an acetic acid (CH3COOH) of unknown concentration is titrated with 0.100 M NaOH. After 10.0 mL of the base solution has been added, the pH in the titration flask is 5.30. What was the concentration of the original acetic acid solution? (Ka(CH3COOH) = 1.8 x 10-5)
Calculate the pH of a 0.100 M sodium benzoate, NaC6H5CO2, solution. The Ka of acetic acid...
Calculate the pH of a 0.100 M sodium benzoate, NaC6H5CO2, solution. The Ka of acetic acid is 6.3 x 10-5
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT