Gas Laws & Stoichiometry

 

Topics:

Previous Topic   (The Mole)

 

Ideal Gas Law

 

 

Chemical reactions also involve substances in the gas phase, therefore to determine the number of moles of a gaseous compound the ideal gas equation is used

 

                        PV = nRT                    where P → pressure in kPa

                                                                      V → volume in litres (L)

                                                                      n → number of moles

                                                                      T → temperature in Kelvin (K)

                                                                      R → universal gas constant; 8.314 kPa • L• mol–1• K–1

 

-the ideal gas law can be used to determine any one of the variables from the knowledge of the other three

 

Example: 

A bulb with a volume of 225 mL contains 0.580 g of an unknown gaseous compound.
The pressure is measured as 145.60 kPa at a temperature of 25 oC. What is the molar mass of the compound?

 

Solution:

 

Convert 225 mL to L;  225 mL = 0.225 L

Convert 25 oC to Kelvin; 25 oC = 273 + 25= 298 K

 

Determine the number of moles of gas under the above conditions

 

n =PV = 145.0 kPa x 0.225 L

     RT     298 K x 8.314 kPa•L•mol–1•K–1

           = 1.32 x 10–2mol

 

1.32 x 10–2mol = 0.580 g

              ! mol  = x g

 

x g = 0.580 g x 1 mol  =  44.0 g/mol

        1.32 x10–2mol

 

Do the following problems:

 

1.      Calculate the volume of 36.0 g of steam at 115 oC and 110.0 kPa of pressure.

2.      If 28 g of N2, nitrogen gas, occupy 22.4 L at STP calculate the volume of 7 g of N2 at 0 oC and 202.6 kPa of pressure.

3.      If 6 g of a gas occupy 20 L at 40 oC and 303.9 kPa of pressure, find its molar mass.

4.      A tire with a volume of 3.60 L contains 0.367 moles of air at a pressure of 252 kPa. What is the temperature, in oC, of the air in the tire.

 

Stoichiometric Calculations
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- stoichiometric calculations make use of balanced equations to calculate quantities of reactants and products

 

Example: 

Given the following balanced equation

 

            3Cu(s)  +  8HNO3(aq) →  3Cu(NO3)2(aq)  +  2NO(g)  +  4H2O(l)

 

determine the volume of 2.50 mol/L nitric acid that would be required to react with 15.5 g of copper.

 

Solutions

 

Use the balanced equations and proceed to write mole ratio, mass ratio and volume ratio.

 

                        3Cu(s)  +  8HNO3(aq) →  3Cu(NO3)2(aq)  +  2NO(g)  +  4H2O(l)

 

mole ratio:       3               8                         3                      2                4

 

mass ratio:     190.65 g    504.16 g        562.71 g             60.02 g       72.08 g

 

vol. ratio                                                                           44.8 L

(STP)

 

            convert 15.5 g of copper to moles

            moles of Cu = 15.5 g x 63.55 g/mol = 0.244 moles

           

            Determine the number of moles of HNO3

            from the equation we see that 3 moles Cu → 8 mole HNO3

                                                          0.244 moles Cu →x moles HNO3

 

                                                            x = 8 moles HNO3 x 0.244 moles Cu  = 0.651 moles HNO3

                                                                                3 moles Cu

 

            Determine the volume of acid, HNO3

 

            Use the formula c = n/v

 

            Find volume; v = n/c

                                   = 0.651 moles/2.50mol/L

                                    = 0.260 L or 260 mL

 

Example:

Given the balanced equation

 

            3Cu(s)  +  8HNO3(aq) →  3Cu(NO3)2(aq)  +  2NO(g)  +  4H2O(l)

 

determine the mass of copper that would be required to produce 4.00 L of nitrogen  monoxide at 102.5 kPa and 22 oC.

 

Solution:

 

Determine the number of moles of NO(g)

Rearrange the ideal gas law equation

 

            n = PV

                  RT

 

            n = _102.5 kPa x 4.00 L             = 0.167 mol NO

                 8.314 kPa L mol−1K−1x 295 K

 

From the balanced equation we see that

 

3 moles Cu → 2 moles NO produced

 

Determine the number of moles of Cu

 

            Moles Cu = 0.167 moles NO x 3 mol Cu = 0.251 mol Cu

                                         2 mol NO

The mass of copper is

 

            Mass of Cu =  0.251 mol Cu x 63.55 g Cu = 16.0 g Cu

                                             1 mol Cu

  

Limiting Reagents

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-         If measured amounts of two substances are reacted then we can calculate the number of moles of

     each reagent used

-         then the balanced equation can be used to determine which material is the limiting reagent

-         the quantity of product will depend on the limiting reagent

 

Sample calculation:

 

Copper reacts with dilute nitric acid according to the equation:

 

            3Cu(s)  +  8HNO3(aq) →  3Cu(NO3)2(aq)  +  2NO(g)  +  4H2O(l)

 

A 200.0 mL sample of 2.00 mol/L nitric acid is added to 30.48 g of copper. Determine which reactant is the limiting reagent. What volume of nitrogen monoxide, measured at 99.5 kPa and 20 oC, would be produced?

 

 solution

 

determine the # of moles of acid

 

moles of HNO3 = 0.2000 L HNO3 x2.00 mol/L HNO3

                          = 0.400 mol HNO3

 

determine the number of moles of copper

 

moles of Cu = 30.48 g χ  63.55 g/mol = 0.4796 mol Cu

 

determine the limiting reagent        

 

from the equation we see that

                                                       3 mol Cu → 8 mol HNO3              using # mole Cu actually have,

                                             0.4796 mol Cu → x mol HNO3              calculate the # of  moles of HNO3                                                                                                        required

 

                                             x mol HNO3 = 8 mol HNO3 x 0.4796 mol Cu

                                                                                 3 mol Cu

                                                                   = 1.279 mol HNO3

 

only 0.400 mol of HNO3 are available, therefore the limiting reagent is HNO3

 

determine the # of moles of NO

 

the amount of limiting reagent 0.400 mol HNO3 determines the amount of NO produced

from the balanced equation we know that the ratio is

 

                                         8 mol HNO3 → 2 mol NO produced

                                          0.400 mol HNO3 → x mol NO

 

                                         x = 2 mol NO x 0.400 mol HNO3

                                                            8 mol HNO3

                                          = 0.100 mol NO

 

determine the volume of NO by using the gas law

 

V = nRT

        P

 

V = 0.100 mol x  8.34 kPa•L•mol–1•K–1 x 293 K

                               99.5 kPa

   = 2.45 L NO

 

Percent yield
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Stoichiometric calculations determine the quantity of product that will form in a reaction ; this represents the maximum possible yield or what is referred to as the theoretical yield

when we carry out a reaction in the laboratory we often obtain less than the theoretical yield; this is the actual yield. The percent yield indicates how much product we produced as a percentage. This is calculated by the expression:

 

                        percent yield =   actual yield      x 100%

                                               theoretical yield

 

Sample calculation:

 

Copper (II) oxide may be reduced to copper by heating it in a stream of hydrogen gas:

 

                        CuO(s)  + H2(g)  →  Cu(s)  +  H2O(g)

 

In an experiment 13.65 g of copper(II) oxide produces 10.75 g of copper metal. What is the percent yield?

 

solution:

 

determine the number of moles of copper(II) oxide

moles of CuO = 13.65 g CuO χ 79.55 g/mol CuO

                       = 0.1716 mol CuO

 

determine the number of moles of Cu

from the equation we see that 1 mol CuO → 1 mol Cu

                                therefore 0.1716 mol CuO → 0.1716 mol Cu

 

determine the possible yield of Cu (theoretical yield)

mass of Cu = 0.1716 mol Cu x 63.55 g/mol

                  = 10.91 g Cu

 

determine the percent yield

percent yield = 10.75 g Cu x 100%  = 98.53%

                         10.91 g Cu

  

PROBLEMS

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1.      “Red lead” is an oxide of lead that has the formula Pb3O4. When boiled with dilute nitric acid, the red lead reacts to produce lead(II) nitrate and lead(IV) oxide:

 

                                    Pb3O4(s)  +  4HNO3(aq)→ 2Pb(NO3)2(aq)  + PbO2(s)  +  2H2O(l)

a)      How many moles of lead(II) nitrate would be produced from the complete reaction of 0.46 mol of red lead with an excess of nitric acid?

b)      What volume of 2.50 mol/L nitric acid would be required to react with 15.85 g of red lead.

c)      What mass of red lead would be required to react to producer 14.75 g of lead ((IV) oxide?

d)      What is the percent yield if 11.25 g of red lead reacts with an excess of nitric acid to produce 3.85 g of lead(IV) oxide?

 

2.      Bromic acid will oxidize sulfur dioxide to sulfuric acid

 

2HBrO3(aq)  +  5SO2(g)  + 4H2O(l)  →  Br2(aq)  +  5H2SO4(aq)

 

In an experiment 0.100 L of 0.200 mol/L bromic acid is reacted with 0.100L of a solution of sulfur dioxide.

a)      What volume of sulfur dioxide measured at 23 oC and 98.7 kPa would neede to be dissolved in the aqueous solution to react will all the bromic acid?

b)      What is the maximum number of moles of sulfuric acid that could be produced?

c)      Calculate the percent yield of sulfuric acid if the concentration of the solution produced was found to be 0.230 mol/L?

 

3.      Aluminum metal can displace silver from a solution of silver nitrate.  In an experiment 0.270 g of aluminum is added to 40.0 mL of 1.00 mol/L silver nitrate solution.

a)      Write a balanced equation for the reaction.

b)      Determine which of the reactants is the limiting reagent.

c)      What is the theoretical yield of silver?

d)      The mass of silver collected in the experiment was 2.98 g. Calculate the percent yield.

 

  

OTHER PROBLEMS:

 

1.      How many atoms are there in one mole of oxygen gas?

 

2.      What is the mass of

a)      238 moles of iron metal

b)      0.0035 moles of Na2HPO4

c)      7.70 x 10−6 moles of helium gas

d)      1.206 x 1024 atoms of N2 gas

 

 3.      How many moles are contained in the following substances

a)      1.50 g of silver metal

b)      a 127 g spool of copper wire

c)      4.00 g sample of argon gas

 

4.      If potassium chlorate is heated gently, the crystals will melt.  Further heating will decompose it to give oxygen gas and potassium chloride.

a)      Write the formula for the reactants and products.

b)      Write the balanced equation for the decomposition of potassium chlorate.

c)      How many grams of potassium chloride are produced when 3.5 moles of potassium chlorate are decomposed?

d)      How many moles of potassium chlorate are needed to give 1.5 moles of oxygen gas.

e)      How many liters of oxygen gas at STP will be produced by decomposing 122.6 g of potassium chlorate?

 

5.      Xenon is a noble gas. One of the first compounds made from xenon was xenon tetrafluoride, XeF4. When 10.0 g of xenon gas react completely  with fluorine to give xenon tetrafluoide, 39.8 kJ of energy are released. The equation for the reaction is:

 

Xe(g)  +  2F2(g)  →  XeF4(s)

 

Using this information calculate how much energy is released when 1 mole of xenon gas reacts.

 

6a) How would you go about preparing 150 mL of 1.5 M barium nitrate solution from solid barium nitrate, Ba(NO3)2?

 b) Determine the molarity of a solution containing 0.15 g of KOH in 25 mL of solution.

 

6.      Solutions of FeBr3 and KOH are mixed and a reddish solid, called a precipitate, Fe(OH)3, is formed.

a)      Write the molecular equation for the above reaction. It must be balanced.

b)      Write the total ionic equation for the reaction.

c)      Write the net ionic equation for the reaction.

 

 

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