This is “The Mole in Chemical Reactions”, section 5.3 from the book Beginning Chemistry (v. 1.0). For details on it (including licensing), click here.
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Consider this balanced chemical equation:
2H_{2} + O_{2} → 2H_{2}OWe interpret this as “two molecules of hydrogen react with one molecule of oxygen to make two molecules of water.” The chemical equation is balanced as long as the coefficients are in the ratio 2:1:2. For instance, this chemical equation is also balanced:
100H_{2} + 50O_{2} → 100H_{2}OThis equation is not conventional—because convention says that we use the lowest ratio of coefficients—but it is balanced. So is this chemical equation:
5,000H_{2} + 2,500O_{2} → 5,000H_{2}OAgain, this is not conventional, but it is still balanced. Suppose we use a much larger number:
12.044 × 10^{23} H_{2} + 6.022 × 10^{23} O_{2} → 12.044 × 10^{23} H_{2}OThese coefficients are also in the ratio of 2:1:2. But these numbers are related to the number of things in a mole: the first and last numbers are two times Avogadro’s number, while the second number is Avogadro’s number. That means that the first and last numbers represent 2 mol, while the middle number is just 1 mol. Well, why not just use the number of moles in balancing the chemical equation?
2H_{2} + O_{2} → 2H_{2}Ois the same balanced chemical equation we started with! What this means is that chemical equations are not just balanced in terms of molecules; they are also balanced in terms of moles. We can just as easily read this chemical equation as “two moles of hydrogen react with one mole of oxygen to make two moles of water.” All balanced chemical reactions are balanced in terms of moles.
Interpret this balanced chemical equation in terms of moles.
P_{4} + 5O_{2} → P_{4}O_{10}Solution
The coefficients represent the number of moles that react, not just molecules. We would speak of this equation as “one mole of molecular phosphorus reacts with five moles of elemental oxygen to make one mole of tetraphosphorus decoxide.”
Test Yourself
Interpret this balanced chemical equation in terms of moles.
N_{2} + 3H_{2} → 2NH_{3}Answer
One mole of elemental nitrogen reacts with three moles of elemental hydrogen to produce two moles of ammonia.
In Chapter 4 "Chemical Reactions and Equations", Section 4.1 "The Chemical Equation", we stated that a chemical equation is simply a recipe for a chemical reaction. As such, chemical equations also give us equivalences—equivalences between the reactants and the products. However, now we understand that these equivalences are expressed in terms of moles. Consider the chemical equation
2H_{2} + O_{2} → 2H_{2}OThis chemical reaction gives us the following equivalences:
2 mol H_{2} ⇔ 1 mol O_{2} ⇔ 2 mol H_{2}OAny two of these quantities can be used to construct a conversion factor that lets us relate the number of moles of one substance to an equivalent number of moles of another substance. If, for example, we want to know how many moles of oxygen will react with 17.6 mol of hydrogen, we construct a conversion factor between 2 mol of H_{2} and 1 mol of O_{2} and use it to convert from moles of one substance to moles of another:
$$\text{17}\text{.6}\overline{){\text{molH}}_{2}}\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}\frac{{\text{1molO}}_{2}}{2\overline{){\text{molH}}_{2}}}=8.80{\text{molO}}_{2}$$Note how the mol H_{2} unit cancels, and mol O_{2} is the new unit introduced. This is an example of a mole-mole calculationA stoichiometry calculation when one starts with moles of one substance and convert to moles of another substance using the balanced chemical equation., when you start with moles of one substance and convert to moles of another substance by using the balanced chemical equation. The example may seem simple because the numbers are small, but numbers won’t always be so simple!
For the balanced chemical equation
2C_{4}H_{10}(g) + 13O_{2} → 8CO_{2}(g) + 10H_{2}O(ℓ)if 154 mol of O_{2} are reacted, how many moles of CO_{2} are produced?
Solution
We are relating an amount of oxygen to an amount of carbon dioxide, so we need the equivalence between these two substances. According to the balanced chemical equation, the equivalence is
13 mol O_{2} ⇔ 8 mol CO_{2}We can use this equivalence to construct the proper conversion factor. We start with what we are given and apply the conversion factor:
$$154\overline{){\text{molO}}_{2}}\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}\frac{{\text{8molCO}}_{2}}{\text{13}\overline{){\text{molO}}_{2}}}=94.8{\text{molCO}}_{2}$$The mol O_{2} unit is in the denominator of the conversion factor so it cancels. Both the 8 and the 13 are exact numbers, so they don’t contribute to the number of significant figures in the final answer.
Test Yourself
Using the above equation, how many moles of H_{2}O are produced when 154 mol of O_{2} react?
Answer
118 mol
It is important to reiterate that balanced chemical equations are balanced in terms of moles. Not grams, kilograms, or liters—but moles. Any stoichiometry problem will likely need to work through the mole unit at some point, especially if you are working with a balanced chemical reaction.
Express in mole terms what this chemical equation means.
CH_{4} + 2O_{2} → CO_{2} + 2H_{2}OExpress in mole terms what this chemical equation means.
Na_{2}CO_{3} + 2HCl → 2NaCl + H_{2}O + CO_{2}How many molecules of each substance are involved in the equation in Exercise 1 if it is interpreted in terms of moles?
How many molecules of each substance are involved in the equation in Exercise 2 if it is interpreted in terms of moles?
For the chemical equation
2C_{2}H_{6} + 7O_{2} → 4CO_{2} + 6H_{2}Owhat equivalences can you write in terms of moles? Use the ⇔ sign.
For the chemical equation
2Al + 3Cl_{2} → 2AlCl_{3}what equivalences can you write in terms of moles? Use the ⇔ sign.
Write the balanced chemical reaction for the combustion of C_{5}H_{12} (the products are CO_{2} and H_{2}O) and determine how many moles of H_{2}O are formed when 5.8 mol of O_{2} are reacted.
Write the balanced chemical reaction for the formation of Fe_{2}(SO_{4})_{3} from Fe_{2}O_{3} and SO_{3} and determine how many moles of Fe_{2}(SO_{4})_{3} are formed when 12.7 mol of SO_{3} are reacted.
For the balanced chemical equation
3Cu(s) + 2NO_{3}^{−}(aq) + 8H^{+}(aq) → 3Cu^{2+}(aq) + 4H_{2}O(ℓ) + 2NO(g)how many moles of Cu^{2+} are formed when 55.7 mol of H^{+} are reacted?
For the balanced chemical equation
Al(s) + 3Ag^{+}(aq) → Al^{3+}(aq) + 3Ag(s)how many moles of Ag are produced when 0.661 mol of Al are reacted?
For the balanced chemical reaction
4NH_{3}(g) + 5O_{2}(g) → 4NO(g) + 6H_{2}O(ℓ)how many moles of H_{2}O are produced when 0.669 mol of NH_{3} react?
For the balanced chemical reaction
4NaOH(aq) + 2S(s) + 3O_{2}(g) → 2Na_{2}SO_{4}(aq) + 2H_{2}O(ℓ)how many moles of Na_{2}SO_{4} are formed when 1.22 mol of O_{2} react?
For the balanced chemical reaction
4KO_{2}(s) + 2CO_{2}(g) → 2K_{2}CO_{3}(s) + 3O_{2}(g)determine the number of moles of both products formed when 6.88 mol of KO_{2} react.
For the balanced chemical reaction
2AlCl_{3} + 3H_{2}O(ℓ) → Al_{2}O_{3} + 6HCl(g)determine the number of moles of both products formed when 0.0552 mol of AlCl_{3} react.
One mole of CH_{4} reacts with 2 mol of O_{2} to make 1 mol of CO_{2} and 2 mol of H_{2}O.
6.022 × 10^{23} molecules of CH_{4}, 1.2044 × 10^{24} molecules of O_{2}, 6.022 × 10^{23} molecules of CO_{2}, and 1.2044 × 10^{24} molecules of H_{2}O
2 mol of C_{2}H_{6} ⇔ 7 mol of O_{2} ⇔ 4 mol of CO_{2} ⇔ 6 mol of H_{2}O
C_{5}H_{12} + 8O_{2} → 5CO_{2} + 6H_{2}O; 4.4 mol
20.9 mol
1.00 mol
3.44 mol of K_{2}CO_{3}; 5.16 mol of O_{2}