12.2: Stoichiometric Calculations
Lesson Objectives
 Calculate the amount in moles of a reactant or product from the mass of another reactant or product. Calculate the mass of a reactant or product from the moles of another reactant or product.
 Calculate the mass of a reactant or product from the mass of another reactant or product.
 Create volume ratios from a balanced chemical equation.
 Use volume ratios and other stoichiometric principles to solve problems involving mass, molar amounts, or volumes of gases.
Check Your Understanding
Recalling Prior Knowledge
 How can a balanced chemical equation be used to construct mole ratios between substances?
 How is a mole ratio used to convert from moles of one reactant or product to moles of another?
The mole ratio is the essence of ideal stoichiometry. The mole ratio tells us the quantitative relationship between reactants and products under ideal conditions, in which all reactants are completely converted into products. In the laboratory, most reactions are not completely ideal. Reactions may not proceed 100% to completion, or a given set of reactants may also undergo side reactions that lead to different products. However, theoretical stoichiometric calculations are important because they allow chemists to know the maximum possible amount of product that can be generated by a reaction from a given amount of each reactant.
Ideal Stoichiometry
Solving any stoichiometric calculation starts with a balanced chemical equation. As we saw in the last lesson, “Mole Ratios,” the coefficients of the balanced equation are the basis for the mole ratio between any pair of reactants and/or products. The following flowchart (Figure below) shows that the conversion from a given substance in moles to moles of an unknown substance involves multiplying by the relevant mole ratio.
This flowchart shows how a mole ratio is used in a stoichiometric conversion problem.
In this lesson, you will expand your understanding of stoichiometry to include the amounts of substances that are measured either by mass or by volume.
MassBased Stoichiometry
While the mole ratio is everpresent in all stoichiometry calculations, amounts of substances in the laboratory are most often measured by mass. Therefore, we need to use molemass calculations in combination with mole ratios to solve several different types of massbased stoichiometry problems.
Mass to Moles Problems
In this type of problem, the mass of one substance is given, usually in grams. This value is then used to determine the amount in moles of another substance that will either react with or be produced from the given substance.


 mass of given → moles of given → moles of unknown

The mass of the given substance is converted into moles by using the molar mass of that substance, which can be calculated from the atomic masses found on a periodic table. Then, the moles of the given substance are converted into moles of the unknown by using the mole ratio from the balanced chemical equation.
Sample Problem 12.3: MassMole Stoichiometry
Tin metal reacts with hydrogen fluoride to produce tin(II) fluoride and hydrogen gas according to the following balanced equation:


 Sn(s) + 2HF(g) → SnF_{2}(s) + H_{2}(g)

How many moles of hydrogen fluoride are required to react completely with 75.0 g of tin?
Step 1: List the known quantities and plan the problem.
Known
 given: 75.0 g Sn
 molar mass of Sn = 118.69 g/mol
 1 mol Sn = 2 mol HF (mole ratio)
Unknown
 mol HF
Use the molar mass of Sn to convert the given mass of Sn to moles. Then use the mole ratio to convert from mol Sn to mol HF. This will be done in a single twostep calculation.


 g Sn → mol Sn → mol HF

Step 2: Solve.



\begin{align*}75.0 \ \text{g Sn} \times \dfrac{1 \ \text{mol Sn}}{118.69 \ \text{g Sn}} \times \dfrac{2 \ \text{mol HF}}{1 \ \text{mol Sn}} = 1.26 \ \text{mol HF}\end{align*}
75.0 g Sn×1 mol Sn118.69 g Sn×2 mol HF1 mol Sn=1.26 mol HF

\begin{align*}75.0 \ \text{g Sn} \times \dfrac{1 \ \text{mol Sn}}{118.69 \ \text{g Sn}} \times \dfrac{2 \ \text{mol HF}}{1 \ \text{mol Sn}} = 1.26 \ \text{mol HF}\end{align*}

Step 3: Think about your result.
The mass of tin is less than one mole, but the 1:2 ratio means that more than one mole of HF is required for the reaction. The answer has three significant figures because the given mass has three significant figures.
 Silver oxide is used in small batteries called button batteries (Figure below). It decomposes upon heating to form silver metal and oxygen gas.


 2Ag_{2}O(s) → 4Ag(s) + O_{2}(g)

How many moles of silver are produced by the decomposition of 4.85 g of Ag_{2}O?
Silver oxide is used in button batteries such as these watch batteries.
Moles to Mass Problems
In this type of problem, the amount of one substance is given in moles. From this, you can determine the mass of another substance that will either react with or be produced from the given substance.


 moles of given → moles of unknown → mass of unknown

The moles of the given substance are first converted into moles of the unknown by using the mole ratio from the balanced chemical equation. Then, the moles of the unknown are converted into a mass (in grams) by using its molar mass, which again can be calculated from the atomic masses given on the periodic table.
Sample Problem 12.4: MoleMass Stoichiometry
Hydrogen sulfide gas burns in oxygen to produce sulfur dioxide and water vapor.


 2H_{2}S(g) + 3O_{2}(g) → 2SO_{2}(g) + 2H_{2}O(g)

What mass of oxygen gas is consumed in a reaction that produces 4.60 moles of SO_{2}?
Step 1: List the known quantities and plan the problem.
Known
 given: 4.60 mol SO_{2}
 2 mol SO_{2} = 3 mol O_{2} (mole ratio)
 molar mass of O_{2} = 32.00 g/mol
Unknown
 mass O_{2} = ? g
Use the mole ratio to convert from mol SO_{2} to mol O_{2}. Then convert mol O_{2} to grams. This will be done in a single twostep calculation.


 mol SO_{2} → mol O_{2} → g O_{2}

Step 2: Solve.



\begin{align*}4.60 \ \text{mol SO}_2 \times \dfrac{3 \ \text{mol O}_2}{2 \ \text{mol SO}_2} \times \dfrac{32.00 \ \text{g O}_2}{1 \ \text{mol O}_2} = 221 \ \text{g O}_2\end{align*}
4.60 mol SO2×3 mol O22 mol SO2×32.00 g O21 mol O2=221 g O2

\begin{align*}4.60 \ \text{mol SO}_2 \times \dfrac{3 \ \text{mol O}_2}{2 \ \text{mol SO}_2} \times \dfrac{32.00 \ \text{g O}_2}{1 \ \text{mol O}_2} = 221 \ \text{g O}_2\end{align*}

Step 3: Think about your result.
According to the mole ratio, 6.90 mol O_{2} is produced, which has a mass of 221 g. The answer has three significant figures because the given value has three significant figures.
 Copper metal reacts with sulfur to form copper(I) sulfide. What mass of copper(I) sulfide is produced by the reaction of 0.528 mol Cu with excess sulfur?
Mass to Mass Problems
Massmass calculations are the most practical of all massbased stoichiometry problems. Moles cannot be measured directly, but masses can be easily measured in the lab for most substances. This type of problem is three steps and is a combination of the two previous types.


 mass of given → moles of given → moles of unknown → mass of unknown

The mass of the given substance is converted into moles by using its molar mass. Then, the moles of the given substance are converted into moles of the unknown by using the mole ratio from the balanced chemical equation. Finally, the moles of the unknown are converted back to a mass by using its molar mass.
Sample Problem 12.5: MassMass Stoichiometry
Ammonium nitrate decomposes to dinitrogen monoxide and water according to the following equation.


 NH_{4}NO_{3}(s) → N_{2}O(g) + 2H_{2}O(l)

In a certain experiment, 45.7 g of ammonium nitrate is decomposed. Find the mass of each of the products formed.
Step 1: List the known quantities and plan the problem.
Known
 given: 45.7 g NH_{4}NO_{3}
 1 mol NH_{4}NO_{3} = 1 mol N_{2}O = 2 mol H_{2}O (mole ratios)
 molar mass of NH_{4}NO_{3} = 80.06 g/mol
 molar mass of N_{2}O = 44.02 g/mol
 molar mass of H_{2}O = 18.02 g/mol
Unknown
 mass N_{2}O = ? g
 mass H_{2}O = ? g
Perform two separate threestep massmass calculations, as shown below.


 g NH_{4}NO_{3} → mol NH_{4}NO_{3} → mol N_{2}O → g N_{2}O



 g NH_{4}NO_{3} → mol NH_{4}NO_{3} → mol H_{2}O → g H_{2}O

Step 2: Solve.



\begin{align*}45.7 \ \text{g NH}_4\text{NO}_3 \times \frac{1 \ \text{mol NH}_4\text{NO}_3}{80.06 \ \text{g NH}_4\text{NO}_3} \times \frac{1 \ \text{mol N}_2\text{O}}{1 \ \text{mol NH}_4\text{NO}_3} \times \frac{44.02 \ \text{g N}_2\text{O}}{1 \ \text{mol N}_2\text{O}} = 25.1 \ \text{g N}_2\text{O}\end{align*}
45.7 g NH4NO3×1 mol NH4NO380.06 g NH4NO3×1 mol N2O1 mol NH4NO3×44.02 g N2O1 mol N2O=25.1 g N2O

\begin{align*}45.7 \ \text{g NH}_4\text{NO}_3 \times \frac{1 \ \text{mol NH}_4\text{NO}_3}{80.06 \ \text{g NH}_4\text{NO}_3} \times \frac{1 \ \text{mol N}_2\text{O}}{1 \ \text{mol NH}_4\text{NO}_3} \times \frac{44.02 \ \text{g N}_2\text{O}}{1 \ \text{mol N}_2\text{O}} = 25.1 \ \text{g N}_2\text{O}\end{align*}




\begin{align*}45.7 \ \text{g NH}_4\text{NO}_3 \times \frac{1 \ \text{mol NH}_4\text{NO}_3}{80.06 \ \text{g NH}_4\text{NO}_3} \times \frac{2 \ \text{mol H}_2\text{O}}{1 \ \text{mol NH}_4\text{NO}_3} \times \frac{18.02 \ \text{g H}_2\text{O}}{1 \ \text{mol H}_2\text{O}} = 20.6 \ \text{g H}_2\text{O}\end{align*}
45.7 g NH4NO3×1 mol NH4NO380.06 g NH4NO3×2 mol H2O1 mol NH4NO3×18.02 g H2O1 mol H2O=20.6 g H2O

\begin{align*}45.7 \ \text{g NH}_4\text{NO}_3 \times \frac{1 \ \text{mol NH}_4\text{NO}_3}{80.06 \ \text{g NH}_4\text{NO}_3} \times \frac{2 \ \text{mol H}_2\text{O}}{1 \ \text{mol NH}_4\text{NO}_3} \times \frac{18.02 \ \text{g H}_2\text{O}}{1 \ \text{mol H}_2\text{O}} = 20.6 \ \text{g H}_2\text{O}\end{align*}

Step 3: Think about your result.
The total mass of the two products is equal to the mass of ammonium nitrate that decomposed, demonstrating the law of conservation of mass. Each answer has three significant figures.
 Solid iron(III) hydroxide reacts with sulfuric acid to produce aqueous iron(III) sulfate and water. What mass of sulfuric acid is needed to completely react with 12.72 g of iron(III) hydroxide? What mass of iron(III) sulfate is produced?
VolumeBased Stoichiometry
As you learned in the chapter on The Mole, Avogadro’s hypothesis states that equal volumes of all gases at the same temperature and pressure contain the same number of gas particles. We also saw that one mole of any gas at standard temperature and pressure (0°C and 1 atm) occupies a volume of 22.4 L. These characteristics make stoichiometry calculations involving gases at STP very straightforward. Consider the following reaction, in which nitrogen and oxygen combine to form nitrogen dioxide.
\begin{align*}
& \text{N}_{2}(g) && + && 2\text{O}_{2}(g) && \rightarrow && 2\text{NO}_{2}(g) \\
& 1 \ \text{molecule} && && 2 \ \text{molecules} && && 2 \ \text{molecules} \\
& 1 \ \text{mol} && && 2 \ \text{mol} && && 2 \ \text{mol} \\
& 1 \ \text{volume} && && 2 \ \text{volumes} && && 2 \ \text{volumes}
\end{align*}
Because of Avogadro’s hypothesis, we know that mole ratios between substances in a gasphase reaction are also volume ratios. The six possible volume ratios for the above equation are:

\begin{align*}\dfrac{1 \ \text{volume N}_2}{2 \ \text{volumes O}_2}\end{align*}
1 volume N22 volumes O2 or \begin{align*}\dfrac{2 \ \text{volumes O}_2}{1 \ \text{volume N}_2}\end{align*}2 volumes O21 volume N2 
\begin{align*}\dfrac{1 \ \text{volume N}_2}{2 \ \text{volumes NO}_2}\end{align*}
1 volume N22 volumes NO2 or \begin{align*}\dfrac{2 \ \text{volumes NO}_2}{1 \ \text{volume N}_2}\end{align*}2 volumes NO21 volume N2 
\begin{align*}\dfrac{2 \ \text{volumes O}_2}{2 \ \text{volumes NO}_2}\end{align*}
2 volumes O22 volumes NO2 or \begin{align*}\dfrac{2 \ \text{volumes NO}_2}{2 \ \text{volumes O}_2}\end{align*}2 volumes NO22 volumes O2
Volume to Volume Problems
The volume ratios above can easily be used when the volume of one gas in a reaction is known and you need to determine the volume of another gas that will either react with or be produced from the first gas. Although pressure and temperature need to be held constant over the course of the reaction for these conversion factors to remain true, the reaction does not need to be run at STP; Avogadro's hypothesis is true regardless of the pressure and temperature being used.
Sample Problem 12.6: VolumeVolume Stoichiometry
The combustion of propane gas produces carbon dioxide and water vapor (see Figure below).


 C_{3}H_{8}(g) + 5O_{2}(g) → 3CO_{2}(g) + 4H_{2}O(g)

A municipal propane tank in Austin, TX. The combustion of propane gas produces carbon dioxide and water vapor.
What volume of oxygen is required to completely combust 0.650 L of propane? What volume of carbon dioxide is produced in the reaction?
Step 1: List the known quantities and plan the problem.
Known
 given: 0.650 L C_{3}H_{8}
 1 volume C_{3}H_{8} = 5 volumes O_{2}
 1 volume C_{3}H_{8} = 3 volumes CO_{2}
Unknown
 volume O_{2} = ? L
 volume CO_{2} = ? L
Step 2: Solve.



\begin{align*}0.650 \ \text{L C}_3\text{H}_8 \times \frac{5 \ \text{L O}_2}{1 \ \text{L C}_3\text{H}_8} = 3.25 \ \text{L O}_2\end{align*}
0.650 L C3H8×5 L O21 L C3H8=3.25 L O2

\begin{align*}0.650 \ \text{L C}_3\text{H}_8 \times \frac{5 \ \text{L O}_2}{1 \ \text{L C}_3\text{H}_8} = 3.25 \ \text{L O}_2\end{align*}




\begin{align*}0.650 \ \text{L C}_3\text{H}_8 \times \frac{3 \ \text{L CO}_2}{1 \ \text{L C}_3\text{H}_8} = 1.95 \ \text{L CO}_2\end{align*}
0.650 L C3H8×3 L CO21 L C3H8=1.95 L CO2

\begin{align*}0.650 \ \text{L C}_3\text{H}_8 \times \frac{3 \ \text{L CO}_2}{1 \ \text{L C}_3\text{H}_8} = 1.95 \ \text{L CO}_2\end{align*}

Step 3: Think about your result.
Because the coefficients on O_{2} and CO_{2} are larger than the one in front of C_{3}H_{8}, the volumes for those two gases are greater. Note that total volume is not necessarily conserved in a reaction because moles are not necessarily conserved. In this reaction, 6 total volumes of reactants become 7 total volumes of products.
 Using the equation for the combustion of propane from Sample Problem 12.6, what volume of water vapor is produced in a reaction that generates 320. mL of carbon dioxide?
Mass to Volume and Volume to Mass Problems
Chemical reactions frequently involve both solid substances whose mass can be measured as well as gases for which measuring the volume is more appropriate. Stoichiometry problems of this type are called either massvolume or volumemass problems.


 mass of given → moles of given → moles of unknown → volume of unknown



 volume of given → moles of given → moles of unknown → mass of unknown

Because both types of problems involve a conversion from either moles of gas to volume or viceversa, we can use the molar volume of 22.4 L/mol as a conversion factor if the reaction is run at STP. In a later chapter (The Behavior of Gases), we will see how to solve this type of problem when other sets of reaction conditions are used.
Sample Problem 12.7: MassVolume Stoichiometry
Aluminum metal reacts rapidly with aqueous sulfuric acid to produce aqueous aluminum sulfate and hydrogen gas.


 2Al(s) + 3H_{2}SO_{4}(aq) → Al_{2}(SO_{4})_{3}(aq) + 3H_{2}(g)

Determine the volume of hydrogen gas produced at STP when a 2.00 g piece of aluminum completely reacts with sulfuric acid.
Step 1: List the known quantities and plan the problem.
Known
 given: 2.00 g Al
 molar mass of Al = 26.98 g/mol
 2 mol Al = 3 mol H_{2}
Unknown
 volume H_{2} = ?
The grams of aluminum will first be converted to moles. Then the mole ratio will be applied to convert to moles of hydrogen gas. Finally, the molar volume of a gas will be used to convert to liters of hydrogen.


 g Al → mol Al → mol H_{2} → L H_{2}

Step 2: Solve.



\begin{align*}2.00 \ \text{g Al} \times \frac{1 \ \text{mol Al}}{26.98 \ \text{g Al}} \times \frac{3 \ \text{mol H}_2}{2 \ \text{mol Al}} \times \frac{22.4 \ \text{L H}_2}{1 \ \text{mol H}_2} = 2.49 \ \text{L H}_2\end{align*}
2.00 g Al×1 mol Al26.98 g Al×3 mol H22 mol Al×22.4 L H21 mol H2=2.49 L H2

\begin{align*}2.00 \ \text{g Al} \times \frac{1 \ \text{mol Al}}{26.98 \ \text{g Al}} \times \frac{3 \ \text{mol H}_2}{2 \ \text{mol Al}} \times \frac{22.4 \ \text{L H}_2}{1 \ \text{mol H}_2} = 2.49 \ \text{L H}_2\end{align*}

Step 3: Think about your result.
The volume result is in liters. For smaller amounts, it may be convenient to convert to milliliters. The answer here has three significant figures. Because the molar volume is a measured quantity of 22.4 L/mol, three is the maximum number of significant figures for this type of problem.
Sample Problem 12.8: VolumeMass Stoichiometry
Calcium oxide is used to trap the sulfur dioxide that is generated in coalburning power plants according to the following reaction:


 2CaO(s) + 2SO_{2}(g) + O_{2}(g) → 2CaSO_{4}(s)

What mass of calcium oxide is required to react completely with 1.2 × 10^{3} L of sulfur dioxide at STP?
Step 1: List the known quantities and plan the problem.
Known
 given: 1.2 × 10^{3} L SO_{2}
 2 mol SO_{2} = 2 mol CaO
 molar mass of CaO = 56.08 g/mol
Unknown
 mass CaO = ? g
The volume of SO_{2} will first be converted to moles. The mole ratio can then be used, and finally, moles of CaO will be converted to grams.


 L SO_{2} → mol SO_{2} → mol CaO → g CaO

Step 2: Solve.
\begin{align*}1.2 \times 10^3 \ \text{L SO}_2 \times \frac{1 \ \text{mol SO}_2}{22.4 \ \text{L SO}_2} \times \frac{2 \ \text{mol CaO}}{2 \ \text{mol SO}_2} \times \frac{56.08 \ \text{g CaO}}{1 \ \text{mol CaO}} = 3.0 \times 10^3 \ \text{g CaO}\end{align*}
Step 3: Think about your result.
The resultant mass could also be reported as 3.0 kg, with two significant figures. Even though the 2:2 mole ratio does not mathematically affect the problem, it is still necessary for unit conversion.
 Sodium azide (NaN_{3}) is a compound that is used in automobile air bags (Figure below). A collision triggers its rapid decomposition into sodium and nitrogen gas, which is the gas that fills the air bag.
 The decomposition of 1.00 g of NaN_{3} produces what volume of N_{2} at STP?
 What mass of NaN_{3} is required to produce 250. L of N_{2} at STP?
The rapid decomposition of sodium azide produces nitrogen gas, which fills an automotive air bag.
Other Stoichiometry
Stoichiometric conversions all involve mole ratios between substances in a balanced chemical equation. Problems that involve mass and/or the volume of a gas are very common and practical. However, a third “arm” of the mole road map could also be part of a problem – the number of representative particles of a substance. Using the equation in Sample Problem 12.7, we could determine the number of formula units of aluminum sulfate produced when 25.0 g of Al reacts.
\begin{align*}25.0 \ \text{g Al} \times \frac{1 \ \text{mol Al}}{26.98 \ \text{g Al}} \times \frac{1 \ \text{mol Al}_2(\text{SO}_4)_3}{2 \ \text{mol Al}} \times \frac{6.02 \times 10^{23} \ \text{form. units Al}_2(\text{SO}_4)_3}{1 \ \text{mol Al}_2(\text{SO}_4)_3} = 2.79 \times 10^{23} \ \text{form. units Al}_2(\text{SO}_4)_3\end{align*}
Problems could potentially arise involving any combination of mass, volume, and number of representative particles. However, since particles of this size cannot actually be counted and stoichiometry is used most often for labbased situations, problems involving the number of particles are seldom encountered in the real world.
Summary of Stoichiometry
The flowchart in Figure below illustrates the types of stoichiometry problems that we have seen in this chapter and that you will most often need to solve. Conversion (b) is always present in any stoichiometry problem, while the use of the conversions represented by (a), (c), (d), and (e) depend on the specific type of problem. Conversion (f) is unique to gaseous volumevolume problems in which the pressure and temperature are held constant.
Flowchart for solving many types of stoichiometry problems.
To learn more about stoichiometric calculations, watch the video lecture at http://www.khanacademy.org/science/chemistry/chemicalreactionsstoichiometry/v/stoichiometry.
An example of how to solve a stoichiometry problem can be seen at http://www.khanacademy.org/science/physics/thermodynamics/v/stoichiometryexampleproblem1.
A second example of how to solve a stoichiometry problem can be seen at http://www.khanacademy.org/science/physics/thermodynamics/v/stoichiometryexampleproblem2.
Lesson Summary
 All stoichiometry problems involve the use of a mole ratio to convert between moles of a given substance and moles of an unknown substance.
 In an ideal stoichiometry problem, the mass of any reactant or product can be calculated from the mass of any other reactant or product if the balanced equation is known.
 Since the molar volume of any gas at STP is constant, gas volumes can also be used in stoichiometric calculations. This allows for multiple types of problems where amounts can be indicated by mass, volume, or number of moles.
Lesson Review Questions
Reviewing Concepts
 How many conversion factors are involved in each of the following stoichiometry problems, where A and B are two components of a chemical reaction, and all volumes are for gases at STP?
 moles of A → mass of B
 volume of A → volume of B
 mass of B → mass of A
 moles of B → volume of A
 Why does a massmass stoichiometry problem require three steps, while a volumevolume stoichiometry problem only requires one step?
Problems
 The doublereplacement reaction that occurs when aqueous solutions of calcium chloride and silver nitrate are combined produces a solution of calcium nitrate and a precipitate of silver chloride.
 How many moles of calcium chloride are needed to react completely with 2.28 moles of silver nitrate?
 If 0.0623 moles of CaCl_{2} reacts completely, how many grams of AgCl are produced?
 In order to produce exactly 1.50 g of AgCl, how many moles of each of the two reactants should be used?
 Silicon dioxide reacts with carbon upon heating to produce silicon carbide (SiC) and carbon monoxide.
 What mass of carbon is required to react completely with 15.70 g of SiO_{2}?
 When 152 g of SiO_{2} reacts with excess carbon, what mass of SiC is produced?
 If 42.2 g of CO were produced by this reaction, what mass of carbon must have reacted?
 Butane (C_{4}H_{10}) combusts according to the following reaction: Assume no change in temperature or pressure for the following questions.
 What volume of O_{2} is needed to combust 425 mL of butane?
 What volume of butane must be combusted to produce 729 L of CO_{2}?
 When 6.20 L of butane is combusted, what volumes of CO_{2} and H_{2}O are produced?
 Dissolving calcium carbonate in hydrochloric acid produces aqueous calcium chloride, carbon dioxide, and water. Assume the reaction takes place at STP.
 What volume of CO_{2} is produced by the reaction of 9.58 g CaCO_{3}?
 If 67.1 L of CO_{2} is produced by this reaction, what mass of HCl must have reacted?
 If 0.812 mol of HCl is used in this reaction, what mass of CaCO_{3} would be consumed? What volume of CO_{2} would be produced?
 Nitroglycerin is an explosive compound that decomposes into multiple gaseous products according to the following reaction:
 What mass of nitrogen gas at STP is produced when 0.314 g of nitroglycerin decomposes?
 What volume of carbon dioxide at STP is produced in a reaction that also produces 2.25 moles of O_{2}?
 What is the total volume of all gases produced at STP by the full decomposition of 10.0 g of nitroglycerin?
 Iron rusts to form iron(III) oxide according to the following equation:
 If 46.2 g of Fe_{2}O_{3} are produced by this reaction, how many Fe atoms reacted?
 What volume of O_{2} at STP is needed to fully react with 8.39 × 10^{24} atoms of iron?
 The complete reaction of 0.916 mol Fe will produce how many formula units of Fe_{2}O_{3}?
 Zinc reacts with hydrochloric acid according to the following equation: In a certain experiment, a 3.77 g sample of impure zinc is reacted with excess hydrochloric acid. If 1.09 L of H_{2} gas is collected at STP, what percentage of the original sample was zinc? Assume that the impurities do not react with HCl.
Further Reading / Supplemental Links
 Stoichiometry, (http://www.wisconline.com/objects/ViewObject.aspx?ID=GCH1504)
 The Chem Collective – Stoichiometry Tutorials – Reaction Stoichiometry, (http://www.chemcollective.org/stoich/reaction_stoi.php)
 Chemical Reaction Stoichiometry, (http://www.chemicalstoichiometry.net/)
 You can view a video lab of acid base stoichiometry at http://www.youtube.com/watch?v=GNNuzEZS8sw.
 The document that goes with this lab is available at http://www.dlt.ncssm.edu/core/Chapter6Stoichiometry/Chapter6Labs/Stoi_AcidBase_HClH2SO4_web_version.doc.
 Here is another online stoichiometry lab. It consists of two videos. Part 1 is at http://www.dlt.ncssm.edu/core/Chapter6Stoichiometry/Chapter6Labs/2_component_system_anal1lg.htm, and part 2 is at http://www.dlt.ncssm.edu/core/Chapter6Stoichiometry/Chapter6Labs/2_component_system_anal2lg.htm.
 The lab document for this is found at http://www.dlt.ncssm.edu/core/Chapter6Stoichiometry/Chapter6Labs/Analysis_of_a_Two_Comp_Sys_web_ver.doc.
 View a microscale stoiciometry lab at http://www.youtube.com/watch?v=vA3o38sFNjY.
 The document for this lab can be found at http://www.dlt.ncssm.edu/core/Chapter6Stoichiometry/Chapter6Labs/Microscale_stoi_web_ver.doc.
 View a stoichiometry experiment involving sodium bicarbonate and hydrochloric acid at http://www.youtube.com/watch?v=vjVrIFScsls.
 The document for this lab is found at http://www.dlt.ncssm.edu/core/Chapter6Stoichiometry/Chapter6Labs/NaHCO3HCl_lab_web_ver.doc.
Points to Consider
This lesson dealt with ideal stoichiometry, where 100% of the reactants were converted to products. In the real world, many chemical reactions do not proceed entirely in this way.
 How can we calculate the amount of products that could be formed in a reaction when two or more reactants are combined in a ratio other than the mole ratio from the balanced equation?
 How can we express the extent to which a set of reactants is converted to products if it is less than 100%?
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