Category Archives: Chemistry homework help



Problem 1 (20 pts total)

Nitrogen is bubbled through a liquid mixture that contains initially equimolar amounts of benzene (B) and toluene (T).  The system pressure is 3 atm and the temperature is 80qC.  The nitrogen flow rate is 10.0 standard liters per second.  The gas leaving the bubbler is saturated with B and T vapors.

  1. Estimate the initial rates (in mol/min) at which B and T leave the bubbler.

Nitrogen gas is bubbled through the system in which a liquid mixture of equimolar amount of benzene and toluene is mixed at

Temp= 80ºC

P = 3 atm

Temperature in Kelvin gives 80+273=353 K

The volumetric flow rate= 10L/min

From ideal gas law PV=nRT


Obtaining volume for benzene

Obtaining vlomue for toluene

Getting the initial rates




  1. How will the mole fractions of B and T in the liquid change with time (increase or decrease or remain the same)? Explain your answer.

The mole fraction of both benzene and toluene will change with time since as the reaction continues, bonding between the atoms of benzene, toluene and nitrogen takes place. This will influence the molarity of the reactants hence triggering the change in mole fraction

  1. How will the mole fractions of B and T in the exiting gas change with time (increase or decrease or remain the same)? Explain your answer.

The mole of the exiting gas will decrease with time since the element in the reactant will keep bonding even though at a lower rate until most of the portion will have formed compounds that cannot leave the bubbler again.


Problem 2 (15 pts total)

A vapor mixture containing 30 mole % of benzene (B) and 70 mole % of toluene (T) at 1 atm is cooled isobarically in a closed container from an initial temperature of 115qC.  Use the Txy diagram (attached below) to answer the following questions:

  1. At what temperature does the first drop of condensate form? What is its composition?


Between 104º to 105º



  1. At one point during the process the system temperature is 100º Determine the mole fraction of B in the vapor and liquid phases and the ratio of total moles in vapor/total moles in liquid at this point.

The vapouor pressure= 0.855

The liquid pressure=0.987

Mole fractions vapour

Mole fraction for liquid

  1. At what temperature does the last bubble of vapor condense? What is its composition?

At the same temperature as the bubble point XB =30% XT=70%

970C <T<990C

Problem 3 (25 pts total) n-butane is converted to isobutane (see reaction below) in a continuous isomerization reactor that operates isothermally at 149qC.  


The feed to the reactor contains 93 mole % of n-butane, 5 mole % of isobutane, and 2 mole % of HCl at 149qC.  A 40% conversion of n-butane is achieved.

  1. Taking a basis of 1 mol of feed gas, calculate the moles of each component of the feed and product mixtures and the extent of reaction (in mol).


  1. Calculate the standard heat of the isomerization reaction (e., heat of reaction) (in kJ/mol). The taking the feed and product species at 25qC as references, prepare an inletoutlet enthalpy table and calculate and fill in the component amount (in mol) and specific enthalpies (in kJ/mol). (Siggia, 1959)

Using Hess’ Law

Ref:  all components at 25ºC 1, atm

Considering that the inlet and outlet are at the same temperature




  1. Calculate the required rate of heat transfer (in kJ) to or from the reactor (state, which it is). Then determine the required heat transfer rate (in kW) for a reactor feed of 325 mol/hr.

EB on open system

Using Heat of Reaction

  1. Use your calculate results to determine the heat of the isomerization reaction at 149q

Problem 4 (40 pts total)

A gas mixtures containing 85 mole % of methane and the balance of oxygen is to be charged into an evacuated well-insulated 10-liter reaction CLOSED (a hint for your calculations of heat management!) vessel at 250 C and 200 kPa.  An electrical coil in the reactor, which delivers heat at a rate of 100 W, will be turned on for 85 seconds and then turned off.  Formaldehyde will be produced in the reaction:

CH4 + O2                             HCHO + H2O

  1. Calculate the maximum pressure that the reactor is likely to have to withstand, assuming that there are no side reactions. If you were ordering the reactor, why would you specify an even greater pressure in your order?  Give at least two reasons.

1 MOLE of gas mixture contains

0.85Moles of Methane (CH4)

0.15 Moles Oxygen (O2)

Temperature 273+25=298 K


HC(298) = -802 KJ/mol


0.85MOL+0.85MOL O2=0.85MOL+0.85MOL

Total moles for oxygen = 0.15+0.15= 0.3

Total moles for methane= 0.85*2= 1.7

From ideal gas equation


It would have the ability to use increased severity to meet hydro treating specifications

It allows for increased operating temperature to meet products specification at lower or higher temperature without melting

  1. Why would heat be added to the feed mixture rather than running the reactor adiabatically?

Running the reactor adiabatically would mean that there is not heat produce hence the reaction being without heat exchange it would also require that the temperature be considered to be varying unlike as it is considered a constant.

  1. Suppose the reaction is run as planned, the reaction products are analyzed chromatographically, and some CO2 is detected. Where did it come from?  If you had taken this CO2 into account, would your calculated pressure in part a) have been larger, or smaller, or you cannot tell without doing the detailed calculations?

During the reaction involving methane and oxygen, the formation of water and Formaldehyde would be more compared to other product. A side reaction due to the presence of head would lead to rapid bonding between carbon and oxygen since this is a reaction that takes place in the presence of the heat. The pressure would be larger since there is an additional reactant in the whole experiment.

(for problem 2)


Siggia, S. (1959). Continuous Analysis of Chemical Process Systems. University of Michigan: Wiley.

Continuous Analysis of Chemical Process Systems
Continuous Analysis of Chemical Process Systems

Spectrophotometeric analysis (changes) of cobalt2+ ions LAB REPORT

The visible light is the portion of the electromagnetic spectrum that can be seen by human eyes.  There are electromagnetic radiation in the range of wavelengths in this perspective that are called the visible light. The longest lengths which the human eyes can respond on this wavelengths ranges between 400 to 700 nanometers which corresponds to the vicinity of 440-770 THz. In the spectrum all colors that can be distinguished by human eyes are not there however there are colors that are termed as the unstructured such as purple or pink while other like magenta are not present because they are made of mixtures of multiple wavelengths. There are colors with single wavelengths which are called spectral colors /pure colors. These wavelengths can pass in large number without attenuated via the earth atmosphere or through the optical windows.

Therefore the wavelength is defined as a measure of the distance between the two identical peaks called crests or between the two high points. The wavelength are used to represent the repeat of a given pattern of the energy in motion like sound and the light.  They have a distinctive formations that plays a vital responsibility in differentiating different types of energies from one to another.  There are specialists in various fields such as scientific and technological professionalisms in identifying the different types of energy. The space in between the repetitions in those waves shows the type of wavelengths that is found on the electromagnetic radiation spectrum. Among them were the waves in the visible light and radio waves of audio range.

The term absorbance is defined as the measure of the ability of a substance or a solution that can enable it to absorb electromagnetic radiation or the light in a specific wavelength. The absorbance is presented by letter (A). The absorbance is equals to the logarithm ratio of the given radiant power of those incident radiation p, to a given power of the transmitted radiation o.  In a given solution, the absorbance can be expressed as the logarithm of ratios of the radiant power of light that can be transmitted over the reference sample of the light that is transmitted through the solution.

Concentration in a chemistry is defined as the abundance of a constituent divide by the volume of the given mixture. There are various many concentrations such as volume concentration, molar concentration, mass concentration and number concentration. Concentration can be found in any kind of chemical mixture. Analytical wave length for quantitative analysis is the wavelength that corresponds to the absorption peaks.

Spectrophotometeric analysis
Spectrophotometeric analysis


In the experiment above the study was to determine the relationship between the color and the wave-lengths the longest and the shortest wave lengths were obtained as follows.

color Short wave length Long wave length
violet 441 473
blue 473 512
green 512 563
yellow 563 575
orange 575 595
red 595 700


From the experiment above, the length of the wavelength which is visible begins at 400-700. The shortest I the blue color ranging from 441-473 when the wave length is increased by 20m while for the highly visible red color ranges from 595-700 when the wave length is equally increased by 20m.  The intensity of the light can be changed by turning the absorbance control by inserting chalk sample as well.  As this process goes the intensity of the light changes. The amount of the energy that can be obtained in the analytical wavelength is that E=hymn use E = hν to get the energy:

E = (6.626 x 10¯34 J s) (5.4545 x 1014 s¯1)

E = 3.614 x 10¯19 J

Now, in order to get the amount of the energy that can be absorbed in one mole of cobalt (II) ions absorbed energy at the wavelength, let’s take the wavelength to be 512 nm.


1) Convert nm to m:

512.0 nm = 512.0 x 10¯9 m = 5.120 x 10¯7 m

The atoms do bond together to form a compound because in doing that they attain lower energies than they possess individual atoms therefore the energy here is less than that of O-H bond.

Preparation of a Co-ordination Compound

Analysis of a Co-ordination Compound

Analysis of a Co-ordination Compound

Spectrophotometric analysis of cobalt.

This section analyses the results of the experiment. The analysis of cobalt is conveniently done by spectrophotometric analysis

Analysis of the visible spectrum of cobalt (II) ion

i.            Prepare solutions of different concentrations by diluting the stock solution with distilled water using a 10 ml pipette.

ii.            Using the wavelength of minimum transmittance found from the visible spectrum of cobalt (II) ion stock solution, measure and record the percentage transmittance and absorbance of each diluted solution.

Analysis of unknown coordination compound of cobalt

i.            Weigh 2.5g of the unknown compound into a dry 50ml beaker.

ii.            Cover the beaker with a watch glass and put it on a hot plate at about 3500C and heat until the sample turns into a liquid, foams and turns blue.

iii.            Remove the beaker and its contents from the hot plate; allow it to cool to room temperature.

iv.            Add 10ml of distilled water and 1 ml of concentrated sulphuric acid to the sample, boil the mixture gently to dissolve the solid. Cool it then to room temperature.

v.            Transfer the contents of the beaker carefully to 100ml volumetric flask, filling it to the mark with distilled water. Mix it thoroughly.

vi.            Measure and record the percentage transmittance and absorbance of the unknown solution at the same wavelength used for the stock solution.

Result interpretation

Percentage cobalt= concentration of Co () (100ml) () ()

Volumetric Analysis of Ammonia Content

Add 50 ml saturated boric acid solution in the 125 ml conical flask; with five drops of the indicator.

Put 1 g of the compound into a beaker and transfer in flask using.

Add Distilled water to give a final volume of about 50 ml.

Add a few boiling chips and several pieces of granulated zinc.

Add 40 ml 6 M NaOH with 50ml distilled water into the flask to recover ammonia.

Gravimetric analysis of chloride


  1. Weigh three sample of the unknown compound of 0.45 to 0.55g, placing them in a 250 ml beaker.
  2. Weigh a second watch glass and carefully transfer the dry crystals onto it. Record the masses and transfer the compound to a weighing bottle. Add about 150 ml of distilled water to the compound and stir to dissolve. Add 1ml of 6M nitric acid to acidify the solution.
  • Slowly add 30 ml of 0.25M silver nitrate, while continuously stirring at room temperature to precipitate the chloride. This is carried in a subdued light to avoid decomposition of silver chloride.
  1. Heat the suspension while stirring until it starts to boil. Stir further for a minute for coagulation of the precipitate. Remove the beaker and contents from heat, cover with a watch glass and leave to stay overnight.
  2. Collect the silver chloride in a cleaned filter crucible making sure that no precipitate is lost.
  3. Wash the residue with 0.01M nitric acid added in small portions.
  • Heat the crucible and contents at 1500C for about an hour, cool it for 30 minutes, weigh it record the mass.

Interpretation of results

The percentage composition of chlode in the unknown compound is

percentage composition of chlode in the unknown compound is


Preparation of a Co-ordination Compound

Preparation of a Co-ordination Compound

Preparation of a Co-ordination Compound


Coordination chemistry is the study of compounds that form between metal ions and other neutral or negatively charged molecules (AP chemistry lab, 2008). This report discusses an experiment to study the Preparation of a Coordination Compound containing cobalt, chlorine and ammonia and the subsequent determination of the formula of the compound. The compound is prepared by the reaction cobalt (ii) chloride 6 hydrate (), ammonium chloride (), concentrated ammonium water (), charcoal and hydrogen peroxide (). Pentaamminechlorocobalt III chloride is synthesized by reacting concentrated ammonia, Cobalt (II) Chloride Hexahydrate, hydrogen peroxide, and concentrated hydrochloric acid together. The objective is to determine the number of atoms using different methods for each ion.



  • Ammonium chloride
  • Distilled water
  • Cobalt (ii) chloride hexahydrate
  • Cobalt solution
  • Powdered charcoal
  • Concentrated ammonia water
  • Hydrogen peroxide
  • Concentrated hydrochloric acid
  • Silver chloride
  • Silver nitrate
  • Concentrated nitric acid
  • Ethyl


  • Thermometer
  • Glass stirring road
  • Clamp and stand
  • 50 ml, 250 ml and 600 ml beaker
  • Erlenmeyer flask
  • Centigram balance
  • Buchner funnel
  • Filter flask
  • Ice bath
  • Fritted glass filter crucible
  • Spatula


Weigh 10g of ammonium chloride on a centigram balance. Add the ammonium chloride to 20ml distilled water in the 50 ml water. Heat the solution until it boils. Add to the boiling solution 15g of cobalt (ii) chloride and record all the masses.

Add hot cobalt solution to the 250 ml Erlenmeyer flask with 1g powdered charcoal. Cool the solution by passing running water. Add 30 ml of concentrated ammonia water to wash any remaining cobalt (ii) solution and transfer it to Erlenmeyer flask.

Cool the flask and its contents to below  in an ice bath. Add carefully 40 ml of hydrogen peroxide to the cobalt (ii) solution in the flask.

Clamp the flask and contents in a 600 ml beaker as water bath. Heat the bath to  maintaing a  more or less variance. Stir the solution with a glass rod, maintaining the same temperature until a pink color appears. Remove the flask and the contents from the water bath once the pink color disappears and cool it to zero degrees.

Filter the crystals formed, retain the residue and discard the filtrate.

Put 125 ml distilled water into a 250 ml beaker and add 5 ml concentrated hydrochloric acid. Heat the solution to boil and transfer the residue to the boiling solution using a spatula.

Upon dissolution filter the hot solution to remove charcoal. Discard the charcoal and save the filtrate.

Add 20 ml of concentrated hydrochloric acid to the filtrate; cool the solution in an ice bath while stirring continuously. At  separate the precipitated crystals and wash the products twice with 15 of 60 by volume ethyl alcohol- water mixture followed by 15 ml by twice 95 by volume of the same.

Discard the filtrate, dry the residue using several thicknesses of filter paper in a watch glass.


When the compound is dissolved in water, the free anions go into solution, but the ones in the complex will remain firmly attached to cobalt. They neither will react with a precipitating agent such as Ag+ ion, which tends to form AgCl with any free Cl-ions. All of the NH3 molecules will be in the complex ion, and will not react with added H+ ion, as they would if they were free

The synthesis of the unknown inorganic compound follows this (unbalanced) equation:

CoCl2· 6 H2O(s) + H2O + NH4Cl(s) + NH3        Cox (NH3)zCly(s)+ H2O(l


Preparation of a Co-ordination Compound