Amounts of Reactants and Products

Amounts of Reactants and Products

** A basic question raised in the chemical laboratory is “How much product will be formed from specific amounts of starting materials (reactants)?” Or in some cases, we might ask the reverse question: “How much starting material must be used to obtain a specific amount of product?”

** To interpret a reaction quantitatively, we need to apply our knowledge of molar masses and the mole concept. Stoichiometry is the quantitative study of reactants and products in a chemical reaction.

** Whether the units given for reactants (or products) are moles, grams, liters (for gases), or some other units, we use moles to calculate the amount of product formed in a reaction. This approach is called the mole method, which means simply that the stoichiometric coefficients in a chemical equation can be interpreted as the number of moles of each substance.

For example, industrially ammonia is synthesized from hydrogen and nitrogen as follows:

** The stoichiometric coefficients show that one molecule of N₂ reacts with three molecules of H2 to form two molecules of NH₃. It follows that the relative numbers of moles are the same as the relative number of molecules:

** Thus, this equation can also be read as “1 mole of N₂ gas combines with 3 moles of H₂ gas to form 2 moles of NH₃ gas.” In stoichiometric calculations, we say that 3 moles of H₂ are equivalent to 2 moles of NH₃, that is,

where the symbol ≈  means “stoichiometrically equivalent to” or simply “equivalent to.” This relationship enables us to write the conversion factors

** Similarly, we have 1 mol N₂ ≈ 2 mol NH3 and 1 mol N₂ ≈ 3 mol H₂.

Let’s consider a simple example in which 6.0 moles of H₂ react completely with N₂ to form NH₃.

** To calculate the amount of NH₃ produced in moles, we use the conversion factor that has H₂ in the denominator and write:

** Now suppose 16.0 g of H₂ react completely with N₂ to form NH₃. How many grams of NH₃ will be formed? To do this calculation, we note that the link between H₂ and NH₃ is the mole ratio from the balanced equation. So we need to first convert grams of H₂ to moles of H₂, then to moles of NH₃, and finally to grams of NH₃. The conversion steps are:

(1) we convert 16.0 g of H₂ to number of moles of H₂, using the molar mass of H₂ as the conversion factor:

(2) we calculate the number of moles of NH₃ produced

(3) we calculate the mass of NH₃ produced in grams using the molar mass of NH₃ as the conversion factor:

** These three separate calculations can be combined in a single step as follows:

** Similarly, we can calculate the mass in grams of N₂ consumed in this reaction. The conversion steps are:

By using the relationship 1 mol N₂ ≈ 3 mol H₂ , we write

** The following Figure shows the steps involved in stoichiometric calculations using the mole method.

First: convert the quantity of reactant A (in grams or other units) to number of moles.  

Next: use the mole ratio in the balanced equation to calculate the number of moles of product B formed.

Finally: convert moles of product to grams of product.

Figure (1)

Solved problems

Problem (1): The food we eat is degraded, or broken down, in our bodies to provide energy for growth and function. A general overall equation for this very complex process represents the degradation of glucose (C₆H₁₂O₆) to carbon dioxide (CO₂) and water (H₂O):

If 968 g of C₆H₁₂O₆ is consumed by a person over a certain period, what is the mass of CO₂ produced?

Strategy:

 Looking at the balanced equation, how do we compare the amount of C₆H₁₂O₆ and CO₂? We can compare them based on the mole ratio from the balanced equation. Starting with grams of C₆H₁₂O₆, how do we convert to moles of C₆H₁₂O₆?

Once moles of CO₂ are determined using the mole ratio from the balanced equation, how do we convert to grams of CO₂?

Solution

 We follow the preceding steps and Figure (1).

Step 1: The balanced equation is given in the problem.

Step 2: To convert grams of C₆H₁₂O₆ to moles of C₆H₁₂O₆, we write

Step 3: From the mole ratio, we see that 1 mol C₆H₁₂O₆ ≈ 6 mol CO₂. Therefore, the

number of moles of CO₂ formed is

Step 4: Finally, the number of grams of CO₂ formed is given by

After some practice, we can combine the conversion steps

into one equation:

Check

Does the answer seem reasonable? Should the mass of CO₂ produced be larger than the mass of C₆H₁₂O₆ reacted, even though the molar mass of CO₂ is considerably less than the molar mass of C₆H₁₂O₆? What is the mole ratio between CO₂ and C₆H₁₂O₆?

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Problem (2): All alkali metals react with water to produce hydrogen gas and the corresponding alkali metal hydroxide. A typical reaction is that between lithium and water:

How many grams of Li are needed to produce 7.79 g of H₂?

Strategy:

 The question asks for number of grams of reactant (Li) to form a specific amount of product (H₂). Therefore, we need to reverse the steps shown in Figure (1).

From the equation we see that 2 mol Li ≈ 1 mol H₂.

Solution:

The conversion steps are

Combining these steps into one equation, we write

Check:

 There are roughly 4 moles of H2 in 7.79 g H₂, so we need 8 moles of Li. From the approximate molar mass of Li (7 g), does the answer seem reasonable?

Reference: Organic chemistry / T.W. Graham Solomons , Craig B.Fryhle , Scott A.snyder , / ( eleventh edition) / 2014.