Stoichiometry and calculations based on stoichiometry


Stoichiometry and calculations based on stoichiometry

Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction. The word ‘Stoichiometry’ is derived from two Greek words – stoicheion (meaning element) and metron (meaning measure). Stoichiometry is an integral part of chemistry that involves the relationship between products and reactants in a chemical reaction. Stoichiometry of chemical reactions indicate the number of moles of reactants and products involved in a balanced chemical reaction and the coefficients involved are called as stoichiometric coefficients.

Zn + 2HCl  ———> ZnCl2 + H2

  1   :     2           :              1     :   1      (Stoichiometric ratio)

N2 + 3H2   ———>     2NH3

   1  : 3            :                  2                (Stoichiometric ratio)

2H2 + O2      ———>  2H2O

   2   :   1         :                 2                 (Stoichiometric ratio)

According to law of conservation of mass;

Mass of H2 + Mass of O2 = Mass of H2O

We cannot find mass of any product or reactant if only mass of one either any reactant or product is given by using law of conservation of mass but we can find it by Stoichiometry of chemical reactions. Almost all stoichiometric problems can be solved in just four simple steps:

  • Balance the equation.
  • Convert units of a given substance to moles.
  • Using the mole ratio, calculate the moles of substance yielded by the reaction.
  • Convert moles of wanted substance to desired unit.

  2H2 + O2  ———->   2H2O

 (Stoichiometric ratio)   2    :   1         :                 2                

 (Given Mass)                10g

 (Molar mass)                 2g     32g                         18g

 (No. of moles)               5        2.5                          5

Example: Two atoms of sulphur react with three molecules of oxygen to form two molecules of sulphur trioxide.

2S + 3O2   —————>  2SO3

How many moles of sulphur react in this way with 9 moles of O2?

Solution: From the balanced equation it is clear that  3O2 react with 2S.

3O2     ——–>   2S

1O2        —–>  2/3 S

9O2               2/3×9 S = 6S

Therefore 6 moles of S must be present in order to react with 9 moles of oxygen.

Limiting Reagent

The limiting reagent (or limiting reactant) in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. This stops the reaction and no further products are made.  Thus it is a substance that limits the amount of product made during a chemical reaction. In other words the reactant that runs out first is known as the limiting reagent. Thus limiting reagent decides the mass of the product formed and the mass of excess reactant that reacts. There is also an excess reactant, or the amount of an element or substance left over after the reaction stops. In a chemical reaction, reactants that are not used up when the reaction is complete are called excess reagents.

While calculating quantity of the product, it will be determined from the reactant present in limiting quantity.

The maximum amount of product that can be produced is called the theoretical yield i.e the amount of the product in g formed from the limiting reagent.

For finding the limiting reagent and theoretical yield, follow the steps as:

 1. Calculate the moles of each reactant present.

2. Calculate the moles of a product formed from each mole of reactant.

3. Recognize the reactant which gives the smaller number of moles of product. This reactant is the Limiting Reagent

 4. Find the grams of product formed by the Limiting Reagent. This is the theoretical yield.

Example: 4Na+O2      —–>        Na2O

Determine the limiting and excess reagant, if you are provided with 20g of Na and 25g of S.

Solution:

     4Na   +   O2         ——>   2Na2O

                4×23

                92g          32g       —–>    124g

   Given:   20g          25g

   According to the balanced chemical equation:                      

   92g of Na reacts with 32g of O2 to produce 124g of Na2O

   Thus 20g of Na react with = 32/92× 20 = 6.95g of O2

Therefore, Na is consumed wholly in the reaction and thus it is the limiting reagent. Since Oxygen is present in excess, thus it is the excess reagent.

Thus mass of oxygen that reacts = 25-6.95 = 18.05g

Example: 3g of H2 react with 29g of O2 to yield H2O. Which is the limiting reagent.

Solution: Express each mass into moles. Since molar mass of H2 = 2.016g and molar mass of O2 = 32g, it follows that

3g of H2 = 3/2.016 = 1.49 mol of H2

and 29g of O2 = 29/32 = 0.9062 mol of O2

The balanced chemical reaction is

2 H2 + O2      ——–>    2 H2O

1.49 mol of H2 require 1.49/2 = 0.745 mol of O2 to yield 1.49 mol of H2O. So Hydrogen is consumed whole and more oxygen is present than required. Thus, Hydrogen is the limiting reagent.

Percentage Yield

 In a few cases the reactions go to completion so that the yield of the reaction is 100%. In most of the cases there are side reactions and the amount of the products formed is less than that based on the calculations. The percent yield is the percent of the product formed based upon the theoretical yield.

Percent Yield = actual yield in g/ theoretical yield x 100 % 

Since the extent of reaction is not necessarily 100%, as a result yield of product formed is less than expected yield (Theoretical yield). The theoretical yield of a reaction can be determined from stoichiometry of chemical reaction.

Example:  What is the % yield of water if 23.2g of water are produced by combining 29g of O2 and 3g of H2

2H2 + O2        ——->     2H2O

Solution: According to balanced equation;

4g H2 require 32g of O2 to yield 36g of H2O

Hydrogen is the limiting reagent.

Thus, 3g of H2 require = 32×3/4 = 24g of O2

Theoretical yield of H2O = 24+3 = 27g

Since Actual yield = 23.2g

Therefore, Percent Yield = actual yield in g/ theoretical yield×100 %

                                         23.2×100/27 = 85.9%

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