What is Gas Stoichiometry?

              Gas stoichiometry uses the ideal gas law to establish the number of moles of a gas in a given volume at any temperature and pressure. It allows us to determine quantities used and formed in chemical reactions.

 

The formula for ideal gas law is:

PV=nRT

where in:

 

• P=pressure

• V=volume

• T=temperature

• n=number of moles of gas

• R=Gas Constant ( 0.0821 L.atm/ K. mol )

 

These are the steps in solving Gas Stoichiometry:

1. Balance the equation for reaction.

2. Convert all amounts to moles.

3. Compare molar amounts using stoichiometry ratios from the balanced equation.

4. Convert the moles into the units required, using the ideal gas law.

 

Example Problems:              

• Ammonia (NH3) gas can be synthesized from nitrogen gas + hydrogen gas.  What volume of ammonia at 450 kPa and 80°C can be obtained from the complete reaction of 7.5 kg hydrogen?

 

First we need a balanced equation:

2Na(s) + 2H2O(l) --> H2(g) + 2NaOH(aq)

 

Ideal Gas Formula:

PV=nRT                                                 Given: P=101.3 kPa, V=20.0 L, T=273 K

        (101.3 kPa)(20.0 L)     =0.893 mol H2   or  mol= 20.0 L (1 mol) =0.893 mol  H2 

  (8.31 kPa•L/K•mol)(273 K)                                           22.4 L 

 

Dimensional Analysis:

g Na=0.839 mol H2  x 2 mol Na x 22. 99 g Na =41.1 g Na

                            1 mol H2      1 mol Na

 

• How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?

    4 Al    +    3 O2    -->   2 Al2O3

 

Ideal Gas Formula:

PV=nRT                                        Given: P=97.3 kPa V=15.0 L  T=21°C=294 K  (97.3 kPa) (15.0 L)=n (8.315dm3 × kPa/mol × K) (294K) n=0.597 mol O2

 

Dimensional Analysis:

0.597 mol O2 x 2 mol Al2O3 x 101. 96 Al2O3 =40. 6 g Al2O3

                    3 mol O2      1 mol Al2O3

 

 

What are Limiting and Excess Reagents?

Limiting Reagent - the reagent that runs out first in a chemical reaction, thus determining the amount of product produced. The reaction will stop when all of the limiting reagent is consumed.

 

Excess Reagent - the reagent that there is a quantity of left over after a chemical reaction. The excess reagent remains because there is nothing with which it can react.

 

Steps in finding limiting and excess reagent:

• Determine the balanced chemical equation for the chemical reaction.

• Convert all given information into moles (most likely, through the use of molar mass as a conversion factor).

• Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.

• Use the amount of limiting reactant to calculate the amount of product produced.

• If necessary, calculate how much is left in excess of the non-limiting reagent.  

 

Examples:

• A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen.  Which is the limiting  and excess reagent?

First, we need to create a balanced equation for the reaction:

4 NH3(g) + 5 O2(g)--->4 NO(g) + 6 H2O(g)

a) 2.00 g NH3 x 1 mol NH3 x 4 mol NO x 30.0 g NO=3. 53 g NO (excess reagent)

                    17.0 g NH3   4 mol NH3   1 mol NO

b) 4.00 g O2 x  1 mol O2 x 4 mol NO x 30.0g NO =3. 00 g NO (limiting reagent)

                   32.0 g O2   5 mol O2    1 mol NO

 

• 2NO + O2 ----> 2 NO2

8 mol NO ------> mol NO2

7 mol O2 ------> mol NO2

a) 8 mol NO=X mol NO2=8 mol NO2 (limiting reagent)

   2 mol O2    2 mol NO2

 

b) 7 mol O2=X mol NO2=14 mol NO2 (excess reagent)

     mol O2    2 mol NO2

 

 

What is the Percent Yield?

Formula:

Percent Yield=    Actual Yield        x 100 %

                       Theoretical Yield

 

Theoretical yield

• The maximum amount of product calculated using the balanced equation.

Actual yield

• The amount of product obtained when the reaction takes place.

Percent yield 

• The ratio of actual yield to theoretical yield.