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Mole Mass Relationship in a Chemical Reaction

A balanced chemical equation provides quantitative information regarding the consumption of reactants and production of products. The numbers which precede the chemical symbols and which balance the equation (with the understanding that if no number appears, it is equal to unity) are called the stoichiometric coefficients (or numbers) and are proportional to the number of molecules or the amounts of the constituents that change during the reaction. For a general reaction,

n 1A + n 2B n 3C+ n 4D

n 1, n 2, n 3 and n 4 are the stoichiometric coefficients for the species A, B, C and D, respectively. The changes in the number of molecules or the amounts of constituents during the course of reaction will be related to each other through the expression  




N stands for the change in number of molecules and Δ n for the change in amount of substance. The negative and positive signs in the above expressions represent, respectively, the decrease and increase in the number of molecules or the amount of the substances. Taking a specific example of the reaction

2H2(g) + O2(g) 2H2O(g)

we can write

In general, for the above reaction, we can state that
  1. 2x molecules of H2(g), combine with 1x molecules O2(g) to give 2x molecules of H2O(g), where x can have any integral value. For example, 2 molecules of H2 combine with 1 molecule of O2 to give 2 molecules of H2O or 4 molecules of H2 combine with 2 molecules of O2 to give 4 molecules of H2O, and so on.
  2. In terms of molecular masses, 2x 2.016 mau of H2 combine with 1x32.00mau of O2 to give 2x(2.016+16.00) mau of H2O, where x can have integral value and mau stands for the atomic mass unit (=1.660310-27 k kg).
  3. In terms of amount of substance, 2y mol of H2 combine with 1y mol of O2 to give 2y mol of H2O, where y can have any numerical value (need not be only an integer). For example, 2 mol of H2 combine with 1 mol of O2 to give 2 mol of H2O or 3 mol of H2 combine with 1.5 mol of O2 to give 3 mol of H2O, and so on.

Chemical Arithmetic

In terms of molar masses, 2y2.016 g of H2 (where 2.016 g mol-1 is the molar mass of H2) combine with 1y32.00 g of O2 (where 32.00 g mol-1 is the molar mass of O2) to give 2y18.016 g of H2O (where 18.016 g mol-1 is the molar mass of H2O). Here y can have any numerical value (need not be only integrals).

The above interpretations also hold good for the reactions taking place in solution. If it is assumed that the volume of the solution does not change during the course of a reaction, then the changes in the reactants and products can also be stated in terms of molar concentrations. By definition, molar concentration is equal to the amount of substance present per dm3 of the solution. The molar concentration is represented by the symbol M and its unit mol dm-3 is represented by M (roman type).

A given reaction may be started with any amount of reactants, but the consumption will take in accordance with the stoichiometric coefficients appearing in the balanced chemical reaction. For example, let the reaction
H2SO4+ 2NaOH Na2SO4 + 2H2O
start with 0.5 mol of H2SO4 and 0.7 mol of NaOH. From the chemical reaction given above, we find that 1 mol of H2SO4 will combine with 2 mol of NaOH. But we are provided with only 0.5 mol of H2SO4, which will react with 20.5 mol, i.e. 1 mol of NaOH. But we are provided with only 0.7 mol of NaOH. Therefore, we can conclude that only 0.35 mol out of 0.5 mol of H2SO4 will react and the remaining 0.15 mol will remain unreacted. Accordingly, the amounts of H2SO4 and water formed will be 0.35 mol and 20.35 mol, respectively. The reactant present in a lesser amount than the required one is known as limiting reactant because its amount limits the amount of products.
We now solve a few problems to illustrate the mole-mass relationship in a chemical reaction.

Potassium bromide contains 32.9 mass % of potassium. If 6.40 g of bromine is made to react with 3.60 g of potassium, calculate the mass of potassium, which combines with bromine to form potassium bromide.

From the given percentage of potassium in potassium bromide, we can proceed as follows:
67.1g of bromine will react with 32.9 g of potassium to give 100 g of potassium bromide. Hence, for the given mass of bromine the mass of potassium that can combine is

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