Friday, June 15, 2012

Powerpoint 5 q's and answers:

1. A gas mixture containing oxygen, nitrogen, and carbon dioxide has a total pressure of 32.9 kPa. If Pressure of oxygen gas = 6.6 kPa, and pressure of nitrogen gas = 23.0 kPa, what is the pressure of carbon dioxide.

Given PO2 = 6.6 kPa
          Ptotal = 32.9 kPa
          PN2= 23.0 kPa

Required: PCO2


Analysis: PCO2 = Ptotal - (PO2 + PN2)
Solve:  PCO2 = 32.9 kPa- ( 6.6 kPa  + 23 kPa+ )
                   = 3.3 kPa

* The pressure of carbon dioxide is 3.3 kPa.

Evaluate: The pressure of the carbon dioxide is suppose to be smaller than the total pressure which means the answer is reasonable.


•2. If I place 3 moles of N2 and 4 moles of O2 in a 35 L container at a temperature of 250 C, what will be the pressure of the resulting mixture of gases be?
• 
•
Given : nnitrogen = 3 moles
             noxygen = 4 moles
             V = 35 L
             T = 25˚C or 298 K
Required: Ptotal
 Analysis: Pnitrogen = nRT/ V
                 Poxygen = nRT/ V
Solve: : Pnitrogen = (3 moles x 8.31 (L*kPa)/ (K*mol) x 298 K) / 35 L
                        = 212 kPa
             Poxygen = (4 moles x 8.31 (L*kPa)/ (K*mol) x 298 K) / 35 L
                        = 283 kPa

             Ptotal = 212 kPa + 283 kPa
                      = 495 kPa
           * The resulting pressure of the two gases is 495 kPa.
   Evaluate: Since the partial pressure of oxygen and nitrogen has been added, thus the total pressure should be greater than their partial pressure. 
 



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Mixed Gas Problems:

1.       If 690.0 mL of oxygen is collected over water at 26.0 ˚C and 725 mmHg, what is the dry volume of this oxygen sample at 52.0 ˚C and 106.6 kPa?
                            Given: P1 = 725 mmHg x (101.3kPa/760mmHg)
                                             = 96.6 kPa
                                          V1= 690.0 mL, T1 = 26.0˚C or 299K, T2 = 52.0˚C or 325 K, P2 = 106.6 kPa  
                             Required: V2
                           Analysis: V2 = (V1 x P1 x T2) / (P­2 x T1)
                           Solve: V2 = (690.0 mL x 96.6 kPa x 325 K)/ (106.6 kPa x 299 K)
                                           = 679.6 mL                                          
                            *The dry volume of the oxygen sample at 52.0 ˚C and 106.6 kPa is 679.6 mL.

                            Evaluate: The answer makes sense because according to Boyle’s Law, pressure is inversely proportional to volume, and since the pressure of the gas increases, the resulting volume decreases.
                          
2.       If 400.0 mL of hydrogen gas is collected over water at 18.0˚C and a total pressure of 98.6 kPa, what would be the dry volume of this hydrogen sample at STP?      
                               Given: V1 = 400.0 mL, T1 = 18˚C or 291 K, P1 = 98.6 kPa, STP= 273 K and 101.3 kPa
                               Required: V2
                               Analysis: V2 = (V1 x P1 x T2) / (P­2 x T1)
                               Solve: V2 = (400mL x 98.6 kPa x 273K)/ (101.3 kPa/ 291 K )
                                                = 365.3 mL  
*The dry volume of this hydrogen at STP is 365.3 L

Evaluate: According to Boyle’s law, pressure is inversely proportional to volume, while according to Charle’s Law temperature is directly proportional to volume; and since the pressure increases and the temperature decreases, the volume decreases.

3.       A 6.12 L sample of xenon gas was collected over water at 2.00 X 105 Pa and 80.0˚C.  What would be the gas volume of pure xenon at SATP?
                                   Given: P1= 2.00 X 10­­5 Pa x (1kPa/1 x 103 Pa)
                                                    =200 kPa
                                               V1= 6.12 L,, T1 = 80.0˚C or 353 K,  SATP= pressure of 101.3kPa and 273 K temperature
                                    Required: V2
                                    Analysis: V2 = (V1 x P1 x T2) / (P­2 x T1)
                                    Solve: V2 = (6.12 L x 200 kPa x 273 K)/ (101.3kPa x 353 K)
                                                      = 9.34 L
            *The gas volume of pure xenon at SATP is 9.34 L.

              Evaluate: : According to Boyle’s law, pressure is inversely proportional to volume, while according to Charle’s Law temperature is directly proportional to volume; and since the pressure decreases and the temperature increases, the expected outcome of volume would be higher than the original.
               
4.       How many moles of gas are contained in 890.0 mL at 21.0˚C and 750.0 mmHg?
                              Given: V= 890.0 mL x ( 1L/1000 mL)
                                             =  0.89 L                                      
                                           T= 21.0˚C or 294 K
                                           P= 750 mmHg x (101.3 kPa / 760 mmHg)
                                            =100Kpa
                                            R= 8.31 (L*kPa)/ (K*mol)
                               Required: mole of the given gas (n)
                               Analysis: n= PV/RT
                               Solve: n= (100kPa x 0.89 L) / (8.31 (L*kPa)/ (K*mol) x 294 K)
                                             = 0.04 mol
             * There’s 0.04 mol of gas contained in 890.0 mL at 21.0˚C and 750.0 mmHg.

Evaluate:


5.       What volume will 20.0 g of Argon gas occupy at STP?
                            Given: margon = 20.0g, STP: T = 273 K, P =101.3 kPa, R=  8.31 (L*kPa)/ (K*mol)
                             Required: nargon and volume of gas
                            Analysis: mol of argon has to be calculated first: n­Ar= 20.0g x (1mol/39.9g)
                                                                                                                        =0.5 mol of argon
                                                V = nRT/P
                                               = [0.5 mol x 8.31 (L*kPa)/ (K*mol ) x 273 K] / 101.3 kPa
                                               = 11.2 L
*The 20.0 g of Argon will occupy 11.2 L at STP.

               
6.       What volume would 32.0 g NO2 gas occupy at 3.12 atm and 18.0˚C? 
                          Given: mNO2 = 32.0 g, T= 18.0˚C or 291 K, R= 8.31 (L*kPa)/ (K*mol)
                                        P= 3.12 atm x (101.3kPa/ 1 atm)
                                          =316.1 kPa
                          Required: V
                          Analysis: Calculate mole of NO2 first,
                                           nNO2 = 32.0 g x [1mol/ 14g + (2x16g)]
                                                     =0.7 mol NO2
                                         
                                             V = nRT/P
                                                = (0.7 mol NO2 x 8.31 (L*kPa)/ (K*mol) x 291 K) / 316.1 kPa
                                                = 5.4 L
                                   *The 32.0 g NO2 gas will occupy 5.4 L at 3.12 atm and 18.0˚C.
 


               
7.       Calculate the molar mass of a gas if 35.44 g of the gas stored in a 7.50 L tank exerts a pressure of 60.0 atm at 35.5˚C.
                        Given: mgas = 35.44g, V= 7.50 L, T= 35.5˚C or 308.5 K, R= 8.31 (L*kPa)/ (K*mol)
                                     P= 60.0 atm x (101.3 kPa/ 1 atm)
                                       = 6078 kPa
                         Required: ngas , Mgas
                         Analyisis: Calculate mole of gas first,
                                             n= PV/ RT
                                             = (6078 kPa x 7.50 L)/ (8.31 (L*kPa)/ (K*mol) x 308.5 K)
                                             = 17.8 mol
                         Solve: solve for molar mass using mole
                                            Mgas = 35.44 g / 17.8 mol
                                                    = 2 g/ mol
                                  *The molar mass of the gas stored in a 7.50 L tank with a pressure of  60.0 atm at 35.5˚C is 2 g / mol.

8.       Determine the number of moles of Krypton contained in a 3.25 L tank at 5.88 X 105 Pa and 25.5˚C.  If the gas was oxygen instead of krypton, will the answer be the same?
                          Given: V= 3.25 L, T= 25.5˚C or 298.5 K, R = 8.31 (L*kPa)/ (K*mol)
                                       P = 5.88 x 105 Pa x (1 kPa/ 1x 103 Pa)
                                          = 588 kPa
                         Required: nKr
                                        Analysis: n= PV/ RT
                         Solve: n= (588 kPa x 3.25 L)/ (8.31 (L*kPa)/ (K*mol) x 298.5 K)
                                       = 0.77 mol
                                * The mole of Krypton contained in a 3.25 L tank at 5.88 X 105 Pa and 25.5˚C is 0.77 mol. 
                          Evaluate: If the gas is oxygen instead of krypton , the answer will be the same unless a specific amount of mass is given in the problem. 
                               
9.       Determine the mass of carbon dioxide in a 450.6 mL flask at 1.80 atm and –50.5˚C.  Determine the mass of oxygen that would be present in the same container under the same conditions.
                             Given: V= 450.6 mL x (1L / 1000mL)
                                            = 0.45 L
                                          P= 1.80 atm x (101.3 kPa/ 1 atm)
                                            = 182.3 kPa
                                          T= -50.5˚C  + 273
                                            = 222.5 K
                                          R= 8.31 (L*kPa)/ (K*mol)
                           Required: nCO2
                           Analysis: n= PV/ RT
                           Solve: nCO2 = (182.3 kPa x 0.45 L) / (8.31 (L*kPa)/ (K*mol) x 222.5 K)
                                               = 0.04 mol
                           MCO2 = 0.04 mol x [ 14g + 2(16g)]/ 1 mol
                                     = 18.4 g CO2
                            
                             nO2 = (182.3 kPa x 0.45 L) / (8.31 (L*kPa)/ (K*mol) x 222.5 K)
                                     = 0.04 mol
                             MO2 = 0.04 mol x (2x16g)/ 1 mol
                                      = 1.28 g O2
                                         * The mass of CO2 in a 450.6 mL flask at 1.80 atm and –50.5˚C is 18.4 g CO2 while the mass of oxygen at the same condition is 1.28 g.


10.   At what temperature will 0.654 mol neon gas occupy 12.30 L at 197.5 kPa?

                            Given: n = 0.654 mol, V= 12.30 L, P= 197.5 kPa, R= 8.31 (L*kPa)/ (K*mol)
                             Required: T
                             Analysis: T= PV/ nR
                             Solve: T= (197.5 kPa x 12.30 L)/ [0.654 mol x 8.31 (L*kPa)/ (K*mol)]
                                           =  447 K
                                *The 0.654 mol neon gas that occupies 12.30 L at 197.5 kPa will have a temperature of 447 K.

11.   A container holds three gases: oxygen, carbon dioxide, and helium.  The partial pressures of the three gases are 2.00 atm, 303.9 kPa, and 3040 mm Hg respectively.  What is the total pressure of this sample in kPa, atm, and mmHg?
                             Given: Poxygen = 2 atm x (101.3 kPa/ 1 atm)
                                                     = 202.6 kPa
                                          Pcarbon dioxide = 303.9 kPa
                                          Phelium = 3040 mmHgx (101.3 kPa/ 760 mmHg)
                                                     = 405.2 kPa
                              Required: PTotal
                              Analysis: PTotal = Poxygen + Pcarbon dioxide + Phelium
                                               Solve: Ptotal = 202.6 kPa + 303.9 kPa +405.2 kPa 
                                                     = 911.7 kPa
                                                                        Or
                                                      911.7kPa x (1atm/ 101.3 kPa)
                                                          = 9 atm
                                                                         Or
                                                     911.7kPa x (760 mmHg/ 101.3kPa)
                                                          = 6849 mmHg
                                 * The total pressure of this sample is 911.7 kPa or 9 atm or 6489 mmHg.
               
12.   A tank contains 480.0 g of oxygen and 80.0 g of helium at a total pressure of 7.00 mmHg. 
What are the partial pressures of oxygen and helium                             
Gas Stoichiometry:
1.       Pentane, C5H12(l), burns completely to form carbon dioxide and water.
a) Write the balanced chemical equation for this reaction.
                         C5H12 + 8O2 → 5CO2 + 6H2O
b) What volume of O2(g) at STP is required to produce 70.0 L of CO2(g) at STP?
                       Given: VCO2 = 70.0 L
                       Required: VO2, nO2, nCO2
                       Analysis: solve for mole of carbon dioxide and oxygen first, and then calculate the volume.
                       Solve: nCO2 =70.0L x ( 1mol/ 22.4 L)
                                           = 3.125 mol CO2
                                
                                  nO2 = 3.125 mol CO2 x (8 mol O2 / 5 mol CO2)
                                         =5 mol O2

                                  VO2 = 5 mol x (22.4 L / 1 mol)
                                         = 112 L of oxygen
                         * The volume of oxygen required to produce 70L of CO2 is 112 L.             
c) What mass of H2O(l) is made when the combustion of C5H12(l) gives 106 L of CO2(g) at SATP?
                                   Given: Vcarbon dioxide 
                                   Required: mwater
                                    Analysis: solve for mole of carbon dioxide and water first, and then calculate the mass of water
                               Solve: : nCO2 =106.0L x ( 1mol/ 22.4 L)
                                                      = 4.7 mol
                                           
                                            nH2O = 4.7 mol CO2 x (6mol H2O/ 5 mol CO2)
                                                      = 5.64 mol H2O

                                            mwater = 5.64mol x [(2 x 1) + 16g / mol]
                                                        = 101.52 g of  H2O
          *101.52 g of H2O(l) is made when the combustion of C5H12(l) gives 106 L of CO2(g) at SATP.
2.       Given the equation   C3H8(g)   +   O2(g)  →   CO2(g)   +    H2O(g)
a)      Balance the equation
                                          C3H8(g)   +   5O2(g)  →   3CO2(g)   +    4H2O(g)
              
b)      If 50 g of C3H8 is burned in excess O2   , what volume of CO2   gas can be collected at 30∧C and 90 kPa?
                               Given: mC3H8 = 50g, T= 30∧C  or 303 K, P= 90 kPa                                                     
                              Required: moles of C3H8 and CO2, VCO2
                               Analysis: Solve for the mole of C3H8 and CO2 first, and then calculate the volume of CO2
                              Solve: n C3H8  = 50 g x [ 1 mol/ (3 x 12) + (8 x 1)g]
                                                     = 1.14 mol C3H8
                                          nCO2 = 1.14 mol C3H8 x (3mol CO2 / 1mol C3H8­)
                                                  = 3.42 mol CO2

                                        VCO2 = nRT/ P
                                                = (3.42 mol x 8.31 (L*kPa)/ (K*mol) x 303 K) / 90kPa
                                                = 95.7 L of CO2
                           * 95.7 L of CO2   gas can be collected at 30∧C and 90 kPa if 50 g of C3H8 is burned in excess O2.   
                    
3.       2 ZnS(s) + 3 O2(g) → 2 ZnO(s) + 2 SO2(g)
a) What volume of O2 at SATP is required for the reaction of 1.46 g of ZnS?
                    Given: m ZnS = 1.46 g
                     Required: mole of O­2 and ZnS; volume of O2
                     Analysis: calculate mole of O­2 and ZnS first, then solve for volume of O2
                     Solve: nZnS= 1.46g x ( 1 mol/ (65.4 + 32.1) g)
                                        = 0.015 mol
                                 nO2 = 0.015 mol ZnS x (3 mol O2 / 2 mol Zns)
                                        = 0.023 mol
                                  VO2= 0.023 mol x (22.4 L / 1 mol)
                                       = 0.52 L O2
                            * It requires 0.52 L O2 for the reaction of 1.46 g of ZnS at SATP

c)       What volume of SO2 at SATP will be produced from the reaction in a)?
                     Given: m ZnS = 1.46 g
                     Required: mole of SO­2 and ZnS; volume of O2
                     Analysis: calculate mole of SO­2 and ZnS first, then solve for volume of SO2
                                    Solve:  nZnS= 1.46g x ( 1 mol/ (65.4 + 32.1) g)
                                                         = 0.015 mol
                                  nSO2 = 0.015 mol ZnS x (2 mol SO2 / 2 mol Zns)
                                        = 0.015 mol SO2
                                  VSO2 = 0.015 mol x (22.4 L / 1 mol)
                                          = 0.34 L SO2
                      
                               * It requires 0.34 L SO2 for the reaction of 1.46 g of ZnS at SATP.




4.       Given the equation         (NH4)2SO4(aq)   +   2KOH(aq)   →   2NH3(g)    +   K2SO4(aq)   +   2H2O(l)
a)      Calculate the volume of ammonia gas, measured at 23∧C and 64 kPa, that could be produced from 264 g of ammonium sulfate and 280.0 g of potassium hydroxide.
                                           Given: T = 23∧C or 296 K, P = 64 kPa, mass of (NH4)2SO4 = 264 g, mass of KOH = 280.0g, R=  8.31 (L*kPa)/ (K*mol)
                                           Required: nKOH , n(NH4)2SO4, nammonia gas, Vammonia gas
                                                 Analysis: calculate the moles of KOH, (NH4)2SO4, and NH3 then solve for volume of ammonia
                                           Solve: nKOH = 280 g x [1mol /(39.1 + 16 + 1)g]
                                                                =5 mol KOH
                                                       nNH3 = 5 mol KOH x (2 mol NH3/ 2 mol KOH)
                                                               = 5 mol NH3

                                                        VNH3 = nRT/P
                                                                 = (5 mol x 8.31 (L*kPa)/ (K*mol) x 296 K) / 64 kPa
                                                         = 192 L

                                                       n(NH4)2SO4 = 264 g x [1mol/ ( (2x14)+ (8 x 2) + 32.1 + ( 4 x 16)]
                                                                        = 1.88 mol
                                                           nNH3 = 1.88 mol (NH4)2SO4 x ( 2 mol KOH/ 1 mol (NH4)2SO4)
                                                                     = 3.76 mol NH­3­

                                                                                         VNH3 = nRT/P
                                                                 = (3.76 mol x 8.31 (L*kPa)/ (K*mol) x 296 K) / 64 kPa
                                                         = 144.5 L


                                          *  At 23∧C and 64 kPa, there will be 192 L of NH3 produced in 280 g KOH and 144.5 L of NH3 produced in 264 g of (NH4)2SO4.
                                  
5.       Given the equation         2KClO3(s)    →   2KCl (s)   +   3O2(g)
What volume of a gas can be produced by the decomposition of 122.6 g of potassium chlorate measured under the following conditions?

a)      at STP
                                              Given: m KClO3 = 122. 6g , P = 101.3 kPa, T = 273 K, R=  8.31 (L*kPa)/ (K*mol)
                                                  Required: n KClO3 , nO2, VO2
                                                                         Analysis: Solve the mole of KClO3 and O2 first, then calculate the volume
                                              Solve: nKClO3 = 122. 6g x (1mol/ [39.1 + 35.5 + (3 x 16)]
                                                                    = 1 mol
                                                           nO2  = 1 mol KClO3 x (3 mol O2/2 mol KClO3)
                                                                   = 1.5 mol O2

                                                            VO2 = nRT/ P
                                                                   =(1.5 mol x 8.31 (L*kPa)/ (K*mol) x 273 K) / 101.3 kPa
                                                           = 33.6 L

                                                * At STP, the  volume of a gas can be produced by the decomposition of 122.6 g of potassium chlorate  is 33.6 L.
                                               
                   

b)      at SATP

                                              Given: m KClO3 = 122. 6g , P = 101.3 kPa, T = 298 K, R=  8.31 (L*kPa)/ (K*mol)
                                                  Required: n KClO3 , nO2, VO2
                                                                         Analysis: Solve the mole of KClO3 and O2 first, then calculate the volume
                                              Solve: nKClO3 = 122. 6g x (1mol/ [39.1 + 35.5 + (3 x 16)]
                                                                    = 1 mol
                                                           nO2  = 1 mol KClO3 x (3 mol O2/2 mol KClO3)
                                                                   = 1.5 mol O2

                                                            VO2 = nRT/ P
                                                                   =(1.5 mol x 8.31 (L*kPa)/ (K*mol) x 298 K) / 101.3 kPa
                                                           = 36.7 L

                                                * At STP, the  volume of a gas can be produced by the decomposition of 122.6 g of potassium chlorate  is 36.7 L.

EVALUATE: The volume at STP condition is smaller than at SATP since the temperature at SATP is greater than STP, and according to Charle's Law, temperature is directly proportional to volume, so the greater volume for SATP is acceptable.  
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