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Solution Set 4

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    36882
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    UC Davis CHE 2A: General Chemistry 
    Instructor: Hayashi

    Chem 2A        Chem 2B        Chem 2C

    Exercises
    Worksheets
    Homework
    Unit I: Atomic Theory            Unit II: Chemical Reactions           Unit III: Gases          Unit IV: Electronic Structure and Bonding

    Q1

    Consider the corrosion of iron via the following reaction:

    \[4Fe_{(s)}+3O_{2\; (g)} \rightarrow 2Fe_2O_{3\; (s)}\]

    How much volume of pure oxygen reacts with 6.66 grams of iron completely at STP?

    1024px-RustyChainEdit1.jpg

    Figure: Heavy rust on the links of a chain near the Golden Gate Bridge in San Francisco. from Wikipedia

    Answer

    • 2.00L

    Hint

    Solution

    Calculate how many moles of oxygen react with Fe by stoichiometry:

    • Moles of Fe= 6.66 g (1 mol/55.85g)= 0.119 mol
    • Moles of O2= moles of Fe(3 moles of O2/4 moles of Fe)= 0.0894 moles

    Use the ideal gas law: PV=nRT V= nRT/P

    At STP, we have P = 1 atm, T = 273K, R= 0.08206L atm/K mol

    V= (0.0894 mol)(0.08206 atm/K mol)(273K)/(1 atm)= 2.00L

    Q2

    Ethanol is produced by fermentation of sugars by yeast. If 60.0 ml Ethanol (\(CH_3CH_2OH\)) with the density of 0.789 g/ml reacts with 17.8 L \(O_2\) at 25°C and a pressure of 2.00 atm to give \(CO_{2(g)}\) and \(H_2O\) via a combustion reaction. When the reaction goes to completion, what is the volume of liquid water produced?

    Answer

    • 26.30 ml

    Hint

    Solution

    First, we can write the balanced chemical equation

    \[C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O\]

    We calculate the moles of the reactant and determine the limiting reagent:

    • Moles of Ethanol: \[n=\dfrac{m}{MM}=\dfrac{D \times V}{MM} = \dfrac{60\; ml \times 0.789\; g/ml}{1 \; mol/46.0674 \;g}= 1.02\;mol\]
    • Moles of Oxygen: PV=nRT, n= PV/RT= (2.00 atm)(17.8 L)/(0.08206 L atm/K mol)(25+273)K=1.46mol

    From 1.02 moles of ethanol, we can have moles of water

    \[= (1.02\; mol\; C_2H_5OH) \times \dfrac{3 \;mol\; H_2O}{1\; mol\; C_2H_5OH}=3.06\;mol\; H_2O\]

    From 1.46 moles of oxygen, we can have moles of water

    \[= (1.46\;mol\; of\; O_2) \times \dfrac{3\; mol H_2O}{3\; mol\; C_2H_5OH} = 1.46\; mol\; H_2O\]

    Therefore, oxygen is the limiting reagent in this reaction.

    N= 1.46 moles N=m/MM

    • Mass of the water =N*MM= 1.46 moles*(18.0158 g/1 mol)= 26.30g
    • Volume of the water= M/D= 26.30 g/1.00 g/ml=26.30 ml

    Q3

    56.0 g liquid Nitrogen is injected into a 10.0 L sealed and empty container. If the surrounding temperature is \(37 ^oC\), what is the final pressure in the container? If this container is already filled with \(Ne_{(g)}\) at 770 mmHg, what will be the partial pressure of \(N_2\) and \(Ne\)? What will be the total pressure in the container?

    1024px-Liquidnitrogen.jpg

    Figure: Liquid nitrogen pour from a dewer. from Wikipedia.

    Answer

    P N2=5.09atm; P N2=5.09atm, PNe=1.01 atm, Ptotal=6.10atm.

    Hint

    Solution

    Step 1: Write down your given information:

    P=?

    \[V=10.0\; L\]

    \[m=56.0\; g\]

    R=0.0820574 L*atm*mol-1K-1

    T=370C

    Step 2: Convert as necessary:

    • Moles: \(56.0\;g\; N_2 \times \dfrac{1\;mol}{28.0\;g}=2.00\;mol\; N_2\)
    • Temperature: \(37^oC + 273\;K=310\;K\)

    Step 3: Plug in the variables into the appropriate equation.

    \[P_{N_2}=\dfrac{nRT}{V}\]

    P N2=2.00mol*0.0820574 L*atm*mol-1K-1*310K/10.0L

    \[P_{N_2}=5.09\; atm\]

    Because the pressure of the container before the \(N_2\) was added contained only Ne, that is your partial pressure of Ne. after converting it to atm, you have already answered part of the question!

    PNe=770 mmHg/760 mmHg / atm=1.01 atm

    Find the unknown moles of Ne.

    nNe=PVRT

    nNe=(1.01atm)(10.00L)/(0.08206atmL/molK)(310K)

    nNe=0.397mol

    Step 3: Now that have pressure for Ne, you must find the partial pressure for N2. Use the ideal gas equation.

    \[\dfrac{P_{Ne}V}{n_{Ne}RT}=\dfrac{P_{N_2}V}{n_{N_2}RT}\]

    But because both gases share the same Volume (V) and Temperature (T) and since the Gas Constant (R) is constants, all three terms cancel and can be removed them from the equation.

    \[\dfrac{P_{Ne}}{n_{Ne}}=\dfrac{P_{N_2}}{n_{N_2}}\]

    \[\dfrac{1.01\; atm}{0.397\;mol}= \dfrac{P_{N_2}}{2.00\; mol}\]

    \[P_{N_2}=5.09\; atm\]

    Ptotal= PNe+PN2=1.01 atm+5.09 atm=6.10 atm

    Q4

    A 12.0 L container contains a mixture of carbon dioxide and helium gas. The room temperature is \(40 ^oC\). If the mole fraction of carbon dioxide is one third of the mixture and its partial pressure is 155.0 Torr, how many moles of helium gas in the container?

    Answer

    • 0.192 moles

    Hint

    Solution

    Step 1: Write down your given information:

    Partial pressure of carbon dioxide=155.0 torr

    Mole fraction of carbon dioxide=1/3

    V=12.0L

    Moles is constant

    R=0.0820574 L*atm*mol-1K-1

    T=400C

    Step 2: Convert as necessary:

    Temperature: 400C+273=313K

    Pressure=155.0 torr*0.00132atm/torr=0.205atm

    Step 3: Plug in the variables into the appropriate equation:

    Mole fraction of helium gas 1-1/3=2/3

    Partial pressure of the helium gas is twice as that of carbon dioxide.

    Partial pressure of helium gas 0.205atm*2=0.410 atm

    PV=nRT

    n=PV/RT

    n=0.410 atm*12.0L/(0.08206Latm/Kmol*313K)=0.192 moles

    Q5

    Water gas is a synthesis gas, containing carbon monoxide and hydrogen, which is produced by passing steam over a red-hot coke (carbon). This product is useful but flammable and poisonous.

    \[H_2O_{(g)} + C_{(s)} \rightarrow H_{2(g)}+ CO_{(g)}\]

    A student wants to collect some water gas for further investigation. For safety reason, he collects the products in a 15.0 L sealed container. 36.0 g steam reacts with a piece of \(300 ^oC\) carbon and water gas has the same temperature as carbon. Assuming all water is consumed in the reaction, please calculate the total pressure and partial pressure of hydrogen gas and carbon monoxide.

    Answer

    • \(P_{total}=12.54 \;atm\)
    • \(P_{CO}=6.27\; atm\)
    • \(P_{H_2}=6.27\; atm\)

    Hint

    Solution

    \[36.0\;g\; H_2O \times \left(\dfrac{1\; mol\; H_2O}{18.0\;g \;H_2O}\right) \times \left(\dfrac{1\;mol\; CO}{1\; mol\; H_2O}\right)=2.00\; mol\; \text{CO produced}\]

    \[P_{CO}=\dfrac{nRT}{V}\]

    \[P_{CO}=\dfrac{2.00\;mol \times (0.0820\;L\; atm/(K \;mol)) \times (300 + 273\;K)}{15 \; L} =6.27\; atm\]

    A similar calculation for

    \[P_{H_2}=\dfrac{2.00\; mol \times (0.0820 \; L \; atm/(K \;mol)) \times (300 + 273\; K) }{15 \; L} = 6.27\;atm\]

    \[P_{total}= P_{CO} +P_{H_2} = 6.27\; atm+6.27\; atm =12.54 \;atm\]

    Q6

    A chemist notices that one of many organic solvents is leaking from the hood. The rate of effusion of this organic gas is 2.866 mL/sec while the effusion of oxygen gas is 3.640 mL/sec. What is the molar mass of this organic solvent?

    Answer

    • \(M= 58.1\; g/mol\)

    Hint

    Solution

    Graham's Law states that the rate of effusion of two different gases at the same conditions are inversely proportional to the square roots of their molar masses as given by the following equation:

    \[ \dfrac{\text{Rate of effusion of A}}{\text{Rate of effusion of B}} =\sqrt{\dfrac{M_B}{M_A}}\]

    Assuming A is oxygen gas and B is the organic solvent

    \[\dfrac{3.640\; mL/sec}{2.866\; mL/sec}=\sqrt{\dfrac{M_B}{36.0\;g/mol}}\]

    \[M_B= 58.1\; g/mol\]

    Q7

    If 193 mL of \(O_2\) were collected over water at an atmospheric pressure of 762 mmHg and a temperature of \(23.0^oC\). How many grams of oxygen were collected?

    Hint

    1. Use Dalton's law and the vapor pressure of water at \(23.0^oC\) to correct the pressure to units of atmospheres. \[P_{Totoal} = P_{O_2} +P_{H_2O}\]
    2. Convert the corrected pressure to atmospheres.
    3. Use the ideal gas law to calculate how many moles of oxygen gas \(n_{O_2}\) were produced: \[PV = nRT\]
    4. Use \(n_{O_2}\) and the molecular weight of oxygen to calculate how many grams of oxygen were collected.

    Solution

    1. Use Dalton's law and the vapor pressure of water at 23.0 o C to correct the pressure to units of atmoshperes.
      PT = Poxygen +Pwater
      At 23.0 o C the vapor pressure of water is 21.1 mmHg. (This can be found on a vapor pressure table.)
      762 mmHg = Poxygen + 21.1 mmHg
      Poxygen = 762 mmHg - 21.1 mmHg
      Poxygen =741 mmHg
    2. Convert the corrected pressure to atmospheres.
      (741 mmHg) (1 atm / 760 mmHg) = 0.975 atm
    3. Use the ideal gas law to find out how many moles of gas were produced:
      PV = nRT (remember to put volume in liters and temperature in Kelvin)
      (0.975 atm) (.193 L) = n (.0821 L atm / mol K) (298 K)
      n = (0.975 atm) (.193 L) / (.0821 L atm / mol K) (298 K)
      n = 7.69 X 10-4 mol
    4. Use the number of moles and the molecular weight of oxygen to find out how many grams of oxygen were collected.

      (7.69 X 10-4 mol) (32.0 g / 1 mol) = 2.46 X 10-2 g


    Q8

    Gas from a burning cigarette lighter was collected over water. After combustion of all hydrocarbon fuel in the lighter, the lighter was 128 mg lighter. The volume of the collected gas was 0.060 L. If the temperature and pressure in the room were at \(35^oC\) and \(1.000 atm\), what is the molar mass of the gas and identity of the hydrocarbon fuel? Note: Partial pressure of water vapor at \(35^oC\) is 42.2 mm Hg.

    Answer

    • 55.2 g/mol

    Q9

    Assuming both temperature (\(130^oC\)) and pressure are kept constant during the oxidation of titanium:

    \[Ti_{(s)} + 2 Cl_{2(g)} \rightarrow TiCl_{4(g)}\]

    How many liters of \(TiCl_4\) gas will be produced after 10.0 L of chlorine react with excess titanium?

    Answer

    • 5.0 L

    Q10

    If 60.0 L of nitrogen is formed at (STP) in the decomposition reaction of ammonia gas:

    \[2 NH_{3(g)} \rightarrow N_{2(g)} + 3 H_{2(g)}\]

    How many liters of hydrogen will be also be produced?

    Answer

    • 180 L

    Q11

    Which of the following pieces of information is NOT needed to calculate the root mean square of a gas?

    • gas constant
    • molar mass
    • temperature
    • pressure
    • volume

    Answer

    • Volume and pressure are not needed

    Q12

    Calculate the root mean square speed, \(u_{rms}\) , in m/s of helium at \(30 ^o C\).

    Solution:

    Start by converting the molar mass for helium from g/mol to kg/mol.

    \(M=(4.00\;g/mol)\times\dfrac{1\;kg}{1000\;g}\)

    \(M=4.00\times10^{-3}\;kg/mol\)

    Now, using the equation for urms substitute in the proper values for each variable and perform the calculation.

    \[u_{rms}=\sqrt{\dfrac{3RT}{M}}\]

    \[u_{rms}=\sqrt{\dfrac{3 \times (8.314\; kg\;m^2/s^2*mol*K)(303K)}{4.00 \times 10^{-3}kg/mol}}\]

    \[u_{rms}=1.37 \times 10^3 \;m/s \]

    Q13

    What is the ratio of \(u_{rms}\) values for helium vs. xenon at \(30^oC\). Which is higher and why?

    Solution:

    There are two approaches to solve this problem: the hard way and the easy way

    Hard way:

    The \(u_{rms}\) speed of helium is calculated from the above example.

    First convert the molar mass of xenon from g/mol to kg/mol as we did for helium in example 1

    \(M_{Xe}=(131.3\;g/mol)\times\dfrac{1\;kg}{1000\;g}\)

    \(M=0.1313\;kg/mol\)

    Now, using the equation for the urms, insert the given and known values and solve for the variable of interest.

    \[u_{rms}=\sqrt{\dfrac{3(8.314\; kg\;m^2/s^2*mol*K)(303\;K)}{0.1313\;kg/mol}}\]

    \[u_{rms}=2.40 \times 10^2\;m/s\]

    Compare the two values for xenon and helium and decide which is greater.

    • \(u_{Xe}=2.4 \times 10^2 \;m/s\)
    • \(u_{He}=1.37 \times10^3 \;m/s\)

    So the ratio of RMS speeds is

    \[\dfrac{u_{Xe}}{u_{He}} \approx 0.18\]

    Helium has the higher \(u_{rms}\) speed. This is in according with Graham's Law, because helium atoms are much lighter than xenon atoms.

    Easy Way:

    Since the temperature is the same for both gases, only the square root of the ratio of molar mass is needed to be calculated.

    \[\sqrt{\dfrac{M_{He}}{M_{Xe}}} = \sqrt{\dfrac{4.00 \; g/mol}{131.3\;g/mol }} \approx 0.18\]

    In either approach, helium has a faster RMS speed than xenon and this is due exclusively to its smaller mass.


    Solution Set 4 is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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