# 6.4: Mole-Mole Relationships in Chemical Reactions

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Learning Objectives

• To use a balanced chemical reaction to determine molar relationships between the substances.

Previously, you learned to balance chemical equations by comparing the numbers of each type of atom in the reactants and products. The coefficients in front of the chemical formulas represent the numbers of molecules or formula units (depending on the type of substance). Here, we will extend the meaning of the coefficients in a chemical equation.

Consider the simple chemical equation

$\ce{2H_2 + O_2 → 2H_2O}$

The convention for writing balanced chemical equations is to use the lowest whole-number ratio for the coefficients. However, the equation is balanced as long as the coefficients are in a 2:1:2 ratio. For example, this equation is also balanced if we write it as

$\ce{4H_2 + 2O_2 → 4H_2O}$

The ratio of the coefficients is 4:2:4, which reduces to 2:1:2. The equation is also balanced if we were to write it as

$\ce{22H_2 + 11O_2 → 22H_2O}$

because 22:11:22 also reduces to 2:1:2.

Suppose we want to use larger numbers. Consider the following coefficients:

$12.044 \times 10^{23}\; H_2 + 6.022 \times 10^{23}\; O_2 → 12.044 \times 10^{23}\; H_2O$

These coefficients also have the ratio 2:1:2 (check it and see), so this equation is balanced. But 6.022 × 1023 is 1 mol, while 12.044 × 1023 is 2 mol (and the number is written that way to make this more obvious), so we can simplify this version of the equation by writing it as

$\ce{2 \;mol\; H_2 + 1\; mol\; O_2 → 2 \;mol\; H_2O}$

We can leave out the word mol and not write the 1 coefficient (as is our habit), so the final form of the equation, still balanced, is

$\ce{2H_2 + O_2 → 2H_2O}$

Now we interpret the coefficients as referring to molar amounts, not individual molecules. The lesson? Balanced chemical equations are balanced not only at the molecular level but also in terms of molar amounts of reactants and products. Thus, we can read this reaction as “two moles of hydrogen react with one mole of oxygen to produce two moles of water.”

2 molecules H2 1 molecule O2 2 molecules H2O

2 moles H2 1 mole O2 2 moles H2O

2 x 2.02 g=4.04 g H2 32.0 g O2 2 x 18.02 g=36.04 g H2O

Figure $$\PageIndex{1}$$: This representation of the production of water from oxygen and hydrogen show several ways to interpret the quantitative information of a chemical reaction.

By the same token, the ratios we constructed to describe molecules reaction can also be constructed in terms of moles rather than molecules. For the reaction in which hydrogen and oxygen combine to make water, for example, we can construct the following ratios:

$\mathrm{\dfrac{2\: mol\: H_2}{1\: mol\: O_2}\: or\: \dfrac{1\: mol\: O_2}{2\: mol\: H_2}}$

$\mathrm{\dfrac{2\: mol\: H_2O}{1\: mol\: O_2}\: or\: \dfrac{1\: mol\: O_2}{2\: mol\: H_2O}}$

$\mathrm{\dfrac{2\: mol\: H_2}{2\: mol\: H_2O}\: or\: \dfrac{2\: mol\: H_2O}{2\: mol\: H_2}}$

We can use these ratios to determine what amount of a substance, in moles, will react with or produce a given number of moles of a different substance. The study of the numerical relationships between the reactants and the products in balanced chemical reactions is called stoichiometry. The ratio of coefficients in a balanced chemical equation, used in computations relating amounts of reactants and products is called the stoichiometric factor.

Example $$\PageIndex{1}$$

How many moles of oxygen react with hydrogen to produce 27.6 mol of H2O? The balanced equation is as follows:

$\ce{2H2 + O2 -> 2H2O} \nonumber$

Solution

Because we are dealing with quantities of H2O and O2, we will use the stoichiometric ratio that relates those two substances. Because we are given an amount of H2O and want to determine an amount of O2, we will use the ratio that has H2O in the denominator (so it cancels) and O2 in the numerator (so it is introduced in the answer). Thus,

$$\mathrm{27.6\: mol\: H_2O\times\dfrac{1\: mol\: O_2}{2\: mol\: H_2O}=13.8\: mol\: O_2}$$

To produce 27.6 mol of H2O, 13.8 mol of O2 react.

Exercise $$\PageIndex{1}$$

Using 2H2 + O2 → 2H2O, how many moles of hydrogen react with 3.07 mol of oxygen to produce H2O?

$$\mathrm{3.07\: mol\: O_2\times\dfrac{2\: mol\: H_2}{1\: mol\: O_2}=6.14\: mol\: H_2}$$

## Concept Review Exercise

1. How do we relate molar amounts of substances in chemical reactions?

1. Amounts of substances in chemical reactions are related by their coefficients in the balanced chemical equation.

## Key Takeaway

• The balanced chemical reaction can be used to determine molar relationships between substances.

## Exercises

1. List the molar ratios you can derive from this balanced chemical equation:

NH3 + 2O2 → HNO3 + H2O

2. List the molar ratios you can derive from this balanced chemical equation

2C2H2 + 5O2 → 4CO2 + 2H2O

3. Given the following balanced chemical equation,

6NaOH + 3Cl2 → NaClO3 + 5NaCl + 3H2O

how many moles of NaCl can be formed if 3.77 mol of NaOH were to react?

4. Given the following balanced chemical equation,

C5H12 + 8O2 → 5CO2 + 6H2O

how many moles of H2O can be formed if 0.0652 mol of C5H12 were to react?

5. Balance the following unbalanced equation and determine how many moles of H2O are produced when 1.65 mol of NH3 react.

NH3 + O2 → N2 + H2O

6. Trinitrotoluene [C6H2(NO2)3CH3], also known as TNT, is formed by reacting nitric acid (HNO3) with toluene (C6H5CH3):

HNO3 + C6H5CH3 → C6H2(NO2)3CH3 + H2O

Balance the equation and determine how many moles of TNT are produced when 4.903 mol of HNO3 react.

7. Chemical reactions are balanced in terms of molecules and in terms of moles. Are they balanced in terms of dozens? Defend your answer.

8. Explain how a chemical reaction balanced in terms of moles satisfies the law of conservation of matter.

1. 1 mol NH3:2 mol O2:1 mol HNO3:1 mol H2O

2. 2 mol C2H2:5 mol O2:4 mol CO2:2 mol H2O
1. 3.14 mol

4. 0.3912 mol
1. 4NH3 + 3O2 → 2N2 + 6H2O; 2.48 mol

6. 3HNO3 + C6H5CH3 → C6H2(NO2)3CH3 + 3H2O; 1.634 mol

7. Yes, they are still balanced.

8. A chemical reaction, balanced in terms of moles, contains the same number of atoms of each element, before and after the reaction. This means that all the atoms and its masses are conserved.