14.2: Stoichiometric Calculations
Lesson Objectives
The student will:
 explain the importance of balancing equations before determining mole ratios.
 determine mole ratios in chemical equations.
 calculate the number of moles of any reactant or product from a balanced equation given the number of moles of one reactant or product.
 calculate the mass of any reactant or product from a balanced equation given the mass of one reactant or product.
Vocabulary
 mole ratio
Introduction
Earlier, we explored mole relationships in balanced chemical equations. In this lesson, we will use the mole as a conversion factor to calculate moles of product from a given number of moles of reactant, or we can calculate the number of moles of reactant from a given number of moles of product. This is called a “molemole” calculation. We will also perform “massmass” calculations, which allow you to determine the mass of reactant required to produce a given amount of product, or the mass of product you can obtain from a given mass of reactant.
Mole Ratios
A mole ratio is the relationship between two components of a chemical reaction. For instance, one way we could read the following reaction is that \begin{align*}2\end{align*}



\begin{align*}2 \ \text{H}_{2(g)} + \text{O}_{2(g)} \rightarrow 2 \ \text{H}_2\text{O}_{(l)}\end{align*}
2 H2(g)+O2(g)→2 H2O(l)

\begin{align*}2 \ \text{H}_{2(g)} + \text{O}_{2(g)} \rightarrow 2 \ \text{H}_2\text{O}_{(l)}\end{align*}

The mole ratio of \begin{align*}\text{H}_{2(g)}\end{align*}



\begin{align*} \frac {2 \ \text{mol} \ \text{H}_2} {1 \ \text{mol} \ \text{O}_2} \ \ \ \ \ \text{or} \ \ \ \ \ \frac {1 \ \text{mol} \ \text{O}_2} {2 \ \text{mol} \ \text{H}_2} \end{align*}
2 mol H21 mol O2 or 1 mol O22 mol H2

\begin{align*} \frac {2 \ \text{mol} \ \text{H}_2} {1 \ \text{mol} \ \text{O}_2} \ \ \ \ \ \text{or} \ \ \ \ \ \frac {1 \ \text{mol} \ \text{O}_2} {2 \ \text{mol} \ \text{H}_2} \end{align*}

What is the ratio of hydrogen molecules to water molecules? By examining the balanced chemical equation, we can see that the coefficient in front of the hydrogen is 2, while the coefficient in front of water is also 2. Therefore, the mole ratio can be written as:



\begin{align*} \frac {2 \ \text{mol} \ \text{H}_2} {2 \ \text{mol} \ \text{H}_2\text{O}} \ \ \ \ \ \text{or} \ \ \ \ \ \frac {2 \ \text{mol} \ \text{H}_2\text{O}} {2 \ \text{mol} \ \text{H}_2} \end{align*}
2 mol H22 mol H2O or 2 mol H2O2 mol H2

\begin{align*} \frac {2 \ \text{mol} \ \text{H}_2} {2 \ \text{mol} \ \text{H}_2\text{O}} \ \ \ \ \ \text{or} \ \ \ \ \ \frac {2 \ \text{mol} \ \text{H}_2\text{O}} {2 \ \text{mol} \ \text{H}_2} \end{align*}

Similarly, the ratio of oxygen molecules to water molecules would be:



\begin{align*} \frac {2 \ \text{mol} \ \text{H}_2\text{O}} {1 \ \text{mol} \ \text{O}_2} \ \ \ \ \ \text{or} \ \ \ \ \ \frac {1 \ \text{mol} \ \text{O}_2} {2 \ \text{mol} \ \text{H}_2\text{O}} \end{align*}
2 mol H2O1 mol O2 or 1 mol O22 mol H2O

\begin{align*} \frac {2 \ \text{mol} \ \text{H}_2\text{O}} {1 \ \text{mol} \ \text{O}_2} \ \ \ \ \ \text{or} \ \ \ \ \ \frac {1 \ \text{mol} \ \text{O}_2} {2 \ \text{mol} \ \text{H}_2\text{O}} \end{align*}

In the following example, let’s try finding the mole ratios by first writing a balanced chemical equation from a “chemical sentence.”
Example:
Four moles of solid aluminum are mixed with three moles of gaseous oxygen to produce two moles of solid aluminum oxide. What is the mole ratio of (1) aluminum to oxygen, (2) aluminum to aluminum oxide, and (3) oxygen to aluminum oxide?
Solution:
Balanced chemical equation: \begin{align*}4 \ \text{Al}_{(s)} + 3 \ \text{O}_{2(g)} \rightarrow 2 \ \text{Al}_2\text{O}_{3(s)}\end{align*}
 mole ratio of aluminum to oxygen = \begin{align*} \frac {4 \ \text{mol Al}} {3 \ \text{mol} \ \text{O}_2} \ \text{or} \ \frac {3 \ \text{mol} \ \text{O}_2} {4 \ \text{mol Al}}\end{align*}
4 mol Al3 mol O2 or 3 mol O24 mol Al  mole ratio of aluminum to aluminum oxide = \begin{align*} \frac {4 \ \text{mol Al}} {2 \ \text{mol Al}_2\text{O}_3} \ \text{or} \ \frac {2 \ \text{mol Al}_2\text{O}_3} {4 \ \text{mol Al}}\end{align*}
4 mol Al2 mol Al2O3 or 2 mol Al2O34 mol Al  mole ratio of oxygen to aluminum oxide = \begin{align*} \frac {3 \ \text{mol O}_2} {2 \ \text{mol Al}_2\text{O}_3} \ \text{or} \ \frac {2 \ \text{mol Al}_2\text{O}_3} {3 \ \text{mol O}_2}\end{align*}
3 mol O22 mol Al2O3 or 2 mol Al2O33 mol O2
Example:
Write the balanced chemical equation for the reaction of solid calcium carbide (\begin{align*}\text{CaC}_2\end{align*}
Solution:
Balanced chemical equation: \begin{align*}\text{CaC}_{2(s)} + 2 \ \text{H}_2\text{O}_{(l)} \rightarrow \text{Ca(OH)}_{2(aq)} + \ \text{C}_2\text{H}_{2(g)}\end{align*}
 mole ratio of calcium carbide to water = \begin{align*} \frac {1 \ \text{mol CaC}_2} {2 \ \text{mol H}_2\text{O}} \ \text{or} \ \frac {2 \ \text{mol H}_2O} {1 \ \text{mol CaC}_2}\end{align*}
1 mol CaC22 mol H2O or 2 mol H2O1 mol CaC2  mole ratio of calcium carbide to calcium hydroxide = \begin{align*} \frac {1 \ \text{mol CaC}_2} {1 \ \text{mol Ca(OH)}_2}\end{align*}
1 mol CaC21 mol Ca(OH)2
The correct mole ratios of the reactants and products in a chemical equation are determined by the balanced equation. Therefore, the chemical equation must always be balanced before the mole ratios are used for calculations. Looking at the unbalanced equation for the reaction of phosphorus trihydride with oxygen, it is difficult to guess the correct mole ratio of phosphorus trihydride to oxygen gas.



\begin{align*}\text{PH}_{3(g)} + \ \text{O}_{2(g)} \rightarrow \text{P}_4\text{O}_{10(s)} + \ \text{H}_2\text{O}_{(g)}\end{align*}
PH3(g)+ O2(g)→P4O10(s)+ H2O(g)

\begin{align*}\text{PH}_{3(g)} + \ \text{O}_{2(g)} \rightarrow \text{P}_4\text{O}_{10(s)} + \ \text{H}_2\text{O}_{(g)}\end{align*}

Once the equation is balanced, however, the mole ratio of phophorus trihydride to oxygen gas is apparent.
Balanced chemical equation: \begin{align*}4 \ \text{PH}_{3(g)} + 8 \ \text{O}_{2(g)} \rightarrow \text{P}_4\text{O}_{10(s)} + 6 \ \text{H}_2\text{O}_{(g)}\end{align*}
The mole ratio of phophorus trihydride to oxygen gas, then, is: \begin{align*} \frac {4 \ \text{mol PH}_3} {8 \ \text{mol O}_2}\end{align*}
Keep in mind that before any mathematical calculations are made relating to a chemical equation, the equation must be balanced.
MoleMole Calculations
In the chemistry lab, we rarely work with exactly one mole of a chemical. In order to determine the amount of reagent (reacting chemical) necessary or the amount of product expected for a given reaction, we need to do calculations using mole ratios.
Look at the following equation. If only \begin{align*}0.50 \ \text{moles}\end{align*} of magnesium hydroxide, \begin{align*}\text{Mg(OH)}_2\end{align*}, are present, how many moles of phosphoric acid, \begin{align*}\text{H}_3\text{PO}_4\end{align*}, would be required for the reaction?


 \begin{align*}2 \ \text{H}_3\text{PO}_4 + 3 \ \text{Mg(OH)}_2 \rightarrow \text{Mg}_3(\text{PO}_4)_2 + 6 \ \text{H}_2\text{O}\end{align*}

Step 1: To determine the conversion factor, we want to convert from moles of \begin{align*}\text{Mg(OH)}_2\end{align*} to moles of \begin{align*}\text{H}_3\text{PO}_4\end{align*}. Therefore, the conversion factor is:


 mole ratio = \begin{align*} \frac {2 \ \text{mol H}_3\text{PO}_4} {3 \ \text{mol Mg(OH)}_2}\end{align*}

Note that what we are trying to calculate is in the numerator, while what we know is in the denominator.
Step 2: Use the conversion factor to answer the question.


 \begin{align*}(0.50 \ \text{mol Mg(OH)}_2) \cdot (\frac {2 \ \text{mol H}_3\text{PO}_4} {3 \ \text{mol Mg(OH)}_2}) = 0.33 \ \text{mol H}_3\text{PO}_4\end{align*}

Therefore, if we have \begin{align*}0.50 \ \text{mol}\end{align*} of \begin{align*}\text{Mg(OH)}_2\end{align*}, we would need \begin{align*}0.33 \ \text{mol}\end{align*} of \begin{align*}\text{H}_3\text{PO}_4\end{align*} to react with all of the magnesium hydroxide. Notice if the equation was not balanced, the amount of \begin{align*}\text{H}_3\text{PO}_4\end{align*} required would have been calculated incorrectly. The ratio would have been 1:1, and we would have concluded that \begin{align*}0.5 \ \text{mol}\end{align*} of \begin{align*}\text{H}_3\text{PO}_4\end{align*} were required.
Example:
How many moles of sodium oxide \begin{align*}(\text{Na}_2\text{O})\end{align*} can be formed from \begin{align*}2.36 \ \text{mol}\end{align*} of sodium nitrate \begin{align*}(\text{NaNO}_3)\end{align*} using the balanced chemical equation below?


 \begin{align*}10 \ \text{Na} + 2 \ \text{NaNO}_3 \rightarrow 6 \ \text{Na}_2\text{O} + \ \text{N}_2\text{O}\end{align*}

Solution:


 \begin{align*}(2.36 \ \text{mol NaNO}_3) \cdot ( \frac {6 \ \text{mol Na}_2\text{O}} {2 \ \text{mol NaNO}_3}) = 7.08 \ \text{mol Na}_2\text{O}\end{align*}

Example:
How many moles of sulfur are required to produce \begin{align*}5.42 \ \text{mol}\end{align*} of carbon disulfide, \begin{align*}\text{CS}_2\end{align*}, using the balanced chemical equation below?


 \begin{align*}\text{C} + 2 \ \text{S} \rightarrow \text{CS}_2\end{align*}

Solution:


 \begin{align*}(5.42 \ \text{mol CS}_2) \cdot ( \frac {2 \ \text{mol S}} {1 \ \text{mol CS}_2}) = 10.84 \ \text{mol S}\end{align*}

MassMass Calculations
A massmass calculation would allow you to solve one of the following types of problems:
 Determine the mass of reactant necessary to product a given amount of product
 Determine the mass of product that would be produced from a given amount of reactant
 Determine the mass of reactant necessary to react completely with a second reactant
As was the case for mole ratios, it is important to double check that you are using a balanced chemical equation before attempting any calculations.
Using Proportion to Solve Stoichiometry Problems
All methods for solving stoichiometry problems contain the same four steps.
Step 1: Write and balance the chemical equation.
Step 2: Convert the given quantity to moles.
Step 3: Convert the moles of known to moles of unknown.
Step 4: Convert the moles of unknown to the requested units.
Step 1 has been covered in previous sections. We also just saw how to complete Step 3 in the previous section by using mole ratios. In order to complete the remaining two steps, we simply need to know how to convert between moles and the given or requested units.
In this section, we will be solving “massmass problems,” which means that both the given value and the requested answer will both be in units of mass, usually grams. Note that if some other unit of mass is used, you should convert to grams first, and use that value for further calculations. The conversion factor between grams and moles is the molar mass (g/mol). To find the number of moles in \begin{align*}x\end{align*} grams of a substance, we divide by the molar mass, and to go back from moles to grams, we multiply by the molar mass. This process is best illustrated through examples, so let’s look at some sample problems.
The balanced equation below shows the reaction between hydrogen gas and oxygen gas to produce water. Since the equation is already balanced, Step 1 is already completed. Remember that the coefficients in the balanced equation are true for moles or molecules, but not for grams.


 \begin{align*}2 \ \text{H}_{2(g)} + \ \text{O}_{2(g)} \rightarrow 2 \ \text{H}_2\text{O}_{(l)}\end{align*}

The molar ratio in this equation is two moles of hydrogen react with one mole of oxygen to produce two moles of water. If you were told that you were going to use 2.00 moles of hydrogen in this reaction, you would also know the moles of oxygen required for the reaction and the moles of water that would be produced. It is only slightly more difficult to determine the moles of oxygen required and the moles of water produced if you were told that you will be using 0.50 mole of hydrogen. Since you are using a quarter as much hydrogen, you would need a quarter as much oxygen and produce a quarter as much water. This is because the molar ratios keep the same proportion. If you were to write out a mathematical equation to describe this problem, you would set up the following proportion:


 \begin{align*} \frac {x \ \text{mol O}_2} {0.50 \ \text{mol H}_2} = \frac {1 \ \text{mol O}_2} {2 \ \text{mol H}_2}\end{align*}

The given quantity, 0.50 mole of hydrogen, is already in moles, so Step 2 is also completed. We set up a proportion to help complete Step 3. From the balanced equation, we know that 1 mole of oxygen reacts with 2 moles of hydrogen. Similarly, we want to determine the \begin{align*}x\end{align*} moles of oxygen needed to react with 0.50 moles of hydrogen. The set up proportion would be similar to the one above. We can then solve the proportion by multiplying the denominator from the left side to both sides of the equal sign. In this case, you will find that \begin{align*}x = 0.25 \ \text{moles of O}_2\end{align*}.
Example:
Pentane, \begin{align*}\text{C}_5\text{H}_{12}\end{align*}, reacts with oxygen gas to produce carbon dioxide and water. How many grams of carbon dioxide will be produced by the reaction of 108.0 grams of pentane?
Step 1: Write and balance the equation.


 \begin{align*}\text{C}_5\text{H}_{12} + 8 \ \text{O}_2 \rightarrow 5 \ \text{CO}_2 + 6 \ \text{H}_2\text{O}\end{align*}

Step 2: Convert the given quantity to moles.


 \begin{align*}\frac {108.0 \ \text{g}} {72.0 \ \text{g/mol}} = 1.50 \ \text{mol C}_5\text{H}_{12}\end{align*}

Step 3: Set up and solve the proportion to find moles of unknown.


 \begin{align*} \frac {x \ \text{mol CO}_2} {1.50 \ \text{mol C}_5\text{H}_{12}} = \frac {5 \ \text{mol CO}_2} {1 \ \text{mol C}_5\text{H}_{12}}\end{align*}

Therefore, \begin{align*}x \ \text{mol CO}_2 = 7.50\end{align*}.
Step 4: Convert the unknown moles to requested units (grams).


 \begin{align*}\text{grams CO}_2 = (7.50 \ \text{mol}) \cdot (44.0 \ \text{g/mol}) = 330. \ \text{grams}\end{align*}

Example:
Aluminum hydroxide reacts with sulfuric acid to produce aluminum sulfate and water. How many grams of aluminum hydroxide are necessary to produce 108 grams of water?
Step 1: Write and balance the equation.


 \begin{align*}2 \ \text{Al(OH)}_3 + 3 \ \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 6 \ \text{H}_2\text{O}\end{align*}

Step 2: Convert the given quantity to moles.


 \begin{align*}\frac {108.0 \ \text{g}} {18.0 \ \text{g/mol}} = 6.00 \ \text{mol H}_2\text{O}\end{align*}

Step 3: Set up and solve the proportion to find moles of unknown.


 \begin{align*} \frac {x \ \text{mol Al(OH)}_3} {6.00 \ \text{mol H}_2\text{O}} = \frac {2 \ \text{mol Al(OH)}_3} {6.00 \ \text{mol H}_2\text{O}}\end{align*}

Therefore, \begin{align*}x \ \text{mol Al(OH)}_3 = 2.00\end{align*}.
Step 4: Convert the moles of unknown to grams.


 \begin{align*}\text{grams Al(OH)}_3 = (2.00 \ \text{mol}) \cdot (78.0 \ \text{g/mol}) = 156 \ \text{grams}\end{align*}

Example:
\begin{align*}15.0\end{align*} grams of chlorine gas is bubbled through liquid sulfur to produce liquid disulfur dichloride. How much product is produced in grams?
Step 1: Write and balance the chemical equation.


 \begin{align*}\text{Cl}_{2(g)} + 2 \ \text{S}_{(l)} \rightarrow \text{S}_2\text{Cl}_{2(l)}\end{align*}

Step 2: Convert the given quantity to moles.


 \begin{align*}\frac {15.0 \ \text{g}} {70.9 \ \text{g/mol}} = 0.212 \ \text{mol}\end{align*}

Step 3: Set up and solve the proportion to find moles of unknown.


 \begin{align*}\frac {x \ \text{mol S}_2\text{Cl}_2} {0.212 \ \text{mol Cl}_2} = \frac {1 \ \text{mol S}_2\text{Cl}_2} {1 \ \text{mol Cl}_2}\end{align*}

Therefore, \begin{align*}x \ \text{mol S}_2\text{Cl}_2 = 0.212\end{align*}.
Step 4: Convert the moles of unknown to grams.


 \begin{align*}\text{grams S}_2\text{Cl}_2 = (0.212 \ \text{mol}) \cdot (135 \ \text{g/mol}) = 28.6 \ \text{grams}\end{align*}

Example:
A thermite reaction occurs between elemental aluminum and iron(III) oxide to produce aluminum oxide and elemental iron. The reaction releases enough heat to melt the iron that is produced. If 500. g of iron is produced in the reaction, how much iron(III) oxide was used as reactant?
Step 1: Write and balance the chemical equation.


 \begin{align*}\text{Fe}_2\text{O}_{3(s)} + 2 \ \text{Al}_{(s)} \rightarrow 2 \ \text{Fe}_{(l)} + \ \text{Al}_2\text{O}_{3(s)}\end{align*}

Step 2: Convert the given quantity to moles.


 \begin{align*}\frac {500. \ \text{g}} {55.9 \ \text{g/mol}} = 8.95 \ \text{mol}\end{align*}

Step 3: Set up and solve the proportion to find moles of unknown.


 \begin{align*}\frac {x \ \text{mol Fe}_2\text{O}_3} {8.95 \ \text{mol Fe}} = \frac {1 \ \text{mol Fe}_2\text{O}_3}{2 \ \text{mol Fe}}\end{align*}

Therefore, \begin{align*}x \ \text{mol Fe}_2\text{O}_3 = 4.48\end{align*}.
Step 4: Convert the moles of unknown to grams.


 \begin{align*}\text{grams Fe}_2\text{O}_3 = (4.48 \ \text{mol}) \cdot (160. \ \text{g/mol}) = 717 \ \text{grams}\end{align*}

Example:
Ibuprofen is a common painkiller used by many people around the globe. It has the formula \begin{align*}\text{C}_{13}\text{H}_{18}\text{O}_2\end{align*}. If \begin{align*}200. \ \text{g}\end{align*} of ibuprofen is combusted, how much carbon dioxide is produced?
Step 1: Write and balance the chemical equation.


 \begin{align*}2 \ \text{C}_{13}\text{H}_{18}\text{O}_{2(s)} + 33 \ \text{O}_{2(s)} \rightarrow 26 \ \text{CO}_{2(g)} + 18 \ \text{H}_2\text{O}_{(l)}\end{align*}

Step 2: Convert the given quantity to moles.


 \begin{align*}\frac {200. \ \text{g}} {206 \ \text{g/mol}} = 0.967 \ \text{mol}\end{align*}

Step 3: Set up and solve the proportion to find moles of unknown.


 \begin{align*} \frac {x \ \text{mol CO}_2} {0.967 \ \text{mol C}_{13}\text{H}_{18}\text{O}_2} = \frac {26 \ \text{mol CO}_2} {2 \ \text{mol C}_{13}\text{H}_{18}\text{O}_2}\end{align*}

Therefore, \begin{align*}x \ \text{mol CO}_2 = 12.6\end{align*}.
Step 4: Convert the moles of unknown to grams.


 \begin{align*}\text{grams CO}_2 = (12.6 \ \text{mol}) \cdot (44.0 \ \text{g/mol}) = 554 \ \text{grams}\end{align*}

Example:
If sulfuric acid is mixed with sodium cyanide, the deadly gas hydrogen cyanide is produced. How much sulfuric acid must be reacted to produce \begin{align*}12.5\end{align*} grams of hydrogen cyanide?
Step 1: Write and balance the chemical equation.


 \begin{align*}2 \ \text{NaCN}_{(s)} + \ \text{H}_2\text{SO}_{4(aq)} \rightarrow \text{Na}_2\text{SO}_{4(aq)} + 2 \ \text{HCN}_{(g)}\end{align*}

Step 2: Convert the given quantity to moles.


 \begin{align*}\frac {12.5 \ \text{g}} {27.0 \ \text{g/mol}} = 0.463 \ \text{mol}\end{align*}

Step 3: Set up and solve the proportion to find moles of unknown.


 \begin{align*} \frac {x \ \text{mol H}_2\text{SO}_4} {0.463 \ \text{mol HCN}} = \frac {1 \ \text{mol H}_2\text{SO}_4} {2 \ \text{mol HCN}}\end{align*}

Therefore, \begin{align*}x \ \text{mol H}_2\text{SO}_4 = 0.232\end{align*}.
Step 4: Convert the moles of unknown to grams.


 \begin{align*}\text{grams H}_2\text{SO}_4 = (0.232 \ \text{mol}) \cdot (98.1 \ \text{g/mol}) = 22.7 \ \text{grams}\end{align*}

A blackboard type discussion of stoichiometry (3e) is available at http://www.youtube.com/watch?v=EdZtSSJecJc (9:21).
Using Dimensional Analysis to Solve Stoichiometry Problems
Many chemists prefer to solve stoichiometry problems with a single line of math instead of writing out the multiple steps. This can be done by using dimensional analysis, also called the factorlabel method. Recall that this is simply a method that uses conversion factors to convert from one unit to another. For a review, refer to the section on dimensional analysis in the chapter “Measurement in Chemistry.” In this method, we can follow the cancellation of units to obtain the correct answer.
Let’s return to some of the problems from the previous section and use dimensional analysis to solve them. For instance: \begin{align*}15.0 \ \text{g}\end{align*} of chlorine gas is bubbled through liquid sulfur to produce disulfur dichloride. How much product is produced in grams?
Step 1: As always, the first step is to correctly write and balance the equation:


 \begin{align*}\text{Cl}_{2(g)} + 2 \ \text{S}_{(l)} \rightarrow \text{S}_2\text{Cl}_{2(l)}\end{align*}

Step 2: Identify what is being given (for this question, \begin{align*}15.0 \ \text{g of Cl}_2\end{align*} is the given) and what is asked for \begin{align*}(\text{grams of S}_2\text{Cl}_2)\end{align*}.
Step 3: Next, use the correct factors that allow you to cancel the units you don’t want and get the unit you do want:
Example:
Consider the thermite reaction again. This reaction occurs between elemental aluminum and iron(III) oxide, releasing enough heat to melt the iron that is produced. If \begin{align*}500.0 \ \text{g}\end{align*} of iron is produced in the reaction, how much iron(III) oxide was placed in the original container?
Step 1: Write and balance the equation:


 \begin{align*}\text{Fe}_2\text{O}_{3(s)} + 2 \ \text{Al}_{(s)} \rightarrow 2 \ \text{Fe}_{(l)} + \ \text{Al}_2\text{O}_{3(s)}\end{align*}

Step 2: Determine what is given and what needs to be calculated:


 \begin{align*} \text{given} = 500. \ \text{g of Fe} \ \ \ \ \ \text{calculate = grams of Fe}_2\text{O}_3\end{align*}

Step 3: Setup the dimensional analysis system:


 \begin{align*}500. \text{g Fe} \cdot \frac {1 \ \text{mol Fe}} {55.85 \ \text{g Fe}} \cdot \frac {1 \ \text{mol Fe}_2\text{O}_3} {2 \ \text{mol Fe}} \cdot \frac {159.7 \ \text{g Fe}_2\text{O}_3} {1 \ \text{mol Fe}_2\text{O}_3} = 717 \ \text{g Fe}_2\text{O}_3\end{align*}

Example:
Ibuprofen is a common painkiller used by many people around the globe. It has the formula \begin{align*}\text{C}_{13}\text{H}_{18}\text{O}_2\end{align*}. If \begin{align*}200. \ \text{g}\end{align*} of Ibuprofen is combusted, how much carbon dioxide is produced?
Step 1: Write and balance the equation:


 \begin{align*}2 \ \text{C}_{13}\text{H}_{18}\text{O}_{2(s)} + 33 \ \text{O}_{2(g)} \rightarrow 26 \ \text{CO}_{2(g)} + 9 \ \text{H}_2\text{O}_{(l)}\end{align*}

Step 2: Determine what is given and what needs to be calculated:


 \begin{align*} \text{given} = 200. \text{g of ibuprofen} \ \ \ \ \ \text{calculate = grams of CO}_2\end{align*}

Step 3: Setup the dimensional analysis system:


 \begin{align*}200. \ \text{g C}_{13}\text{H}_{18}\text{O}_2 \cdot \frac {1 \ \text{mol C}_{13}\text{H}_{18}\text{O}_2} {206.3 \ \text{g C}_{13}\text{H}_{18}\text{O}_2} \cdot \frac {26 \ \text{mol CO}_2} {2 \ \text{mol C}_{13}\text{H}_{18}\text{O}_2} \cdot \frac {44.1 \ \text{g CO}_2} {1 \ \text{mol CO}_2} = 555 \ \text{g CO}_2\end{align*}

Example:
If sulfuric acid is mixed with sodium cyanide, the deadly gas hydrogen cyanide is produced. How much sulfuric acid must be placed in the container to produce \begin{align*}12.5 \ \text{g}\end{align*} of hydrogen cyanide?
Step 1: Write and balance the equation:


 \begin{align*}2 \ \text{NaCN}_{(s)} + \ \text{H}_2\text{SO}_{4(aq)} \rightarrow \text{Na}_2\text{SO}_{4(s)} + 2 \ \text{HCN}_{(g)}\end{align*}

Step 2: Determine what is given and what needs to be calculated:


 \begin{align*}\text{given} = 12.5 \ \text{g HCN} \ \ \ \ \ \text{calculate = grams of H}_2\text{SO}_4\end{align*}

Step 3: Setup the dimensional analysis system:


 \begin{align*}12.5 \ \text{g HCN} \cdot \frac {1 \ \text{mol HCN}} {27.0 \ \text{g HCN}} \cdot \frac {1 \ \text{mol H}_2\text{SO}_4} {2 \ \text{mol HCN}} \cdot \frac {98.06 \ \text{g H}_2\text{SO}_4} {1 \ \text{mol H}_2\text{SO}_4} = 22.7 \ \text{g H}_2\text{SO}_4\end{align*}

Lesson Summary
 The coefficients in a balanced chemical equation represent the relative amounts of each substance in the reaction.
 When the moles of one substance in a reaction is known, the coefficients of the balanced equation can be used to determine the moles of all the other substances.
 Massmass calculations can be done using dimensional analysis.
Review Questions
 How many moles of water vapor can be produced from \begin{align*}2\ \mathrm{moles}\end{align*} of ammonia for the following reaction between ammonia and oxygen: \begin{align*}4 \ \text{NH}_{3(g)} + 5 \ \text{O}_{2(g)} \rightarrow 4 \ \text{NO}_{(g)} + 6 \ \text{H}_2\text{O}_{(g)}\end{align*}?
 \begin{align*}3 \ \mathrm{mol}\end{align*}
 \begin{align*}6 \ \mathrm{mol}\end{align*}
 \begin{align*}12 \ \mathrm{mol}\end{align*}
 \begin{align*}24 \ \mathrm{mol}\end{align*}
 How many moles of bismuth(III) oxide can be produced from \begin{align*}0.625 \ \mathrm{mol}\end{align*} of bismuth in the following reaction: \begin{align*}\text{Bi}_{(s)} + \ \text{O}_{2(g)} \rightarrow \text{Bi}_2\text{O}_{3(s)}\end{align*}? (Note: equation may not be balanced.)
 \begin{align*} 0.313 \ \mathrm{mol}\end{align*}
 \begin{align*}0.625 \ \mathrm{mol}\end{align*}
 \begin{align*}1 \ \mathrm{mol}\end{align*}
 \begin{align*}1.25 \ \mathrm{mol}\end{align*}
 \begin{align*}2 \ \mathrm{mol}\end{align*}
 For the following reaction, balance the equation and then determine the mole ratio of moles of \begin{align*}\text{B(OH)}_3\end{align*} to moles of water: \begin{align*}\text{B}_2\text{O}_{3(s)} + \ \text{H}_2\text{O}_{(l)} \rightarrow \text{B(OH)}_{3(s)}\end{align*}.
 \begin{align*}1:1\end{align*}
 \begin{align*}2:3\end{align*}
 \begin{align*}3:2\end{align*}
 None of the above.
 Write the balanced chemical equation for the reactions below and find the indicated molar ratio.
 Gaseous propane \begin{align*}(\text{C}_3\text{H}_8)\end{align*} combusts to form gaseous carbon dioxide and water. Find the molar ratio of \begin{align*}\text{O}_2\end{align*} to \begin{align*}\text{CO}_2\end{align*}.
 Solid lithium reacts with an aqueous solution of aluminum chloride to produce aqueous lithium chloride and solid aluminum. Find the molar ratio of \begin{align*}\text{AlCl}_{3(aq)}\end{align*} to \begin{align*}\text{LiCl}_{(aq)}\end{align*}.
 Gaseous ethane \begin{align*}(\text{C}_2\text{H}_6)\end{align*} combusts to form gaseous carbon dioxide and water. Find the molar ratio of \begin{align*}\text{CO}_{2(g)}\end{align*} to \begin{align*}\text{O}_{2(g)}\end{align*}.
 An aqueous solution of ammonium hydroxide reacts with an aqueous solution of phosphoric acid to produce aqueous ammonium phosphate and water. Find the molar ratio of \begin{align*}\text{H}_3\text{PO}_{4(aq)}\end{align*} to \begin{align*}\text{H}_2\text{O}_{(l)}\end{align*}.
 Solid rubidium reacts with solid phosphorous to produce solid rubidium phosphide. Find the molar ratio of Rb(s) to P(s).
 For the given reaction (unbalanced): \begin{align*}\text{Ca}_3(\text{PO}_4)_2 + \ \text{SiO}_2 + \ \text{C} \rightarrow \text{CaSiO}_3 + \ \text{CO} + \ \text{P}\end{align*}
 how many moles of silicon dioxide are required to react with \begin{align*}0.35 \ \mathrm{mol}\end{align*} of carbon?
 how many moles of calcium phosphate are required to produce \begin{align*}0.45 \ \mathrm{mol}\end{align*} of calcium silicate?
 For the given reaction (unbalanced): \begin{align*}\text{FeS} + \ \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 + \ \text{SO}_2\end{align*}
 how many moles of iron(III) oxide are produced from \begin{align*}1.27 \ \mathrm{mol}\end{align*} of oxygen?
 how many moles of iron(II) sulfide are required to produce \begin{align*}3.18 \ \mathrm{mol}\end{align*} of sulfur dioxide?
 Write the following balanced chemical equation. Ammonia and oxygen are allowed to react in a closed container to form nitrogen and water. All species present in the reaction vessel are in the gas state.
 How many moles of ammonia are required to react with \begin{align*}4.12 \ \mathrm{mol}\end{align*} of oxygen?
 How many moles of nitrogen are produced when \begin{align*}0.98 \ \mathrm{mol}\end{align*} of oxygen are reacted with excess ammonia?
 How many grams of nitric acid will react with \begin{align*}2.00 \ \mathrm{g}\end{align*} of copper(II) sulfide given the following reaction between copper(II) sulfide and nitric acid: \begin{align*}3 \ \text{CuS}_{(s)} + 8 \ \text{HNO}_{3(aq)} \rightarrow 3 \text{Cu(NO}_3)_{2(aq)} + 2 \ \text{NO}_{(g)} + 4 \ \text{H}_2\text{O}_{(l)} + 3 \ \text{S}_{(s)}\end{align*}?
 \begin{align*}0.49 \ \mathrm{g}\end{align*}
 \begin{align*}1.31 \ \mathrm{g}\end{align*}
 \begin{align*}3.52 \ \mathrm{g}\end{align*}
 \begin{align*}16.0 \ \mathrm{g}\end{align*}
 When properly balanced, what mass of iodine is needed to produce \begin{align*}2.5 \ \mathrm{g}\end{align*} of sodium iodide in the following equation: \begin{align*}\text{I}_{2(aq)} + \ \text{Na}_2\text{S}_2\text{O}_{3(aq)} \rightarrow \text{Na}_2\text{S}_4\text{O}_{6(aq)} + \ \text{NaI}_{(aq)}\end{align*}?
 \begin{align*}1.0 \ \mathrm{g}\end{align*}
 \begin{align*}2.1 \ \mathrm{g}\end{align*}
 \begin{align*}2.5 \ \mathrm{g}\end{align*}
 \begin{align*}8.5 \ \mathrm{g}\end{align*}
 Donna was studying the following reaction for a stoichiometry project: \begin{align*}\text{S}_{(s)} + 3 \ \text{F}_{2(g)} \rightarrow \text{SF}_{6(s)}\end{align*} She wondered how much she could obtain if she used \begin{align*}3.5 \ \mathrm{g}\end{align*} of fluorine. What mass of \begin{align*}\text{SF}_6(s)\end{align*} would she obtain from the calculation using this amount of fluorine?
 \begin{align*}3.5\ \mathrm{g}\end{align*}
 \begin{align*}4.5 \ \mathrm{g}\end{align*}
 \begin{align*}10.5\ \mathrm{g}\end{align*}
 \begin{align*}13.4 \ \mathrm{g}\end{align*}
 Aqueous solutions of aluminum sulfate and sodium phosphate are placed in a reaction vessel and allowed to react. The products of the reaction are aqueous sodium sulfate and solid aluminum phosphate.
 Write a balanced chemical equation to represent the above reaction.
 How many grams of sodium phosphate must be added to completely react all of \begin{align*}5.00 \ \mathrm{g}\end{align*} of aluminum sulfate?
 If \begin{align*}3.65 \ \mathrm{g}\end{align*} of sodium phosphate were placed in the container, how many grams of sodium sulfate would be produced?
 For the given reaction (unbalanced): \begin{align*}\text{Ca(NO}_3)_2 + \ \text{Na}_3\text{PO}_4 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + \ \text{NaNO}_3\end{align*}
 how many grams of sodium nitrate are produced from \begin{align*}0.35 \ \mathrm{g}\end{align*} of sodium phosphate?
 how many grams of calcium nitrate are required to produce \begin{align*}5.5\ \mathrm{g}\end{align*} of calcium phosphate?
 For the given reaction (unbalanced): \begin{align*}\text{Na}_2\text{S} + \ \text{Al(NO}_3)_3 \rightarrow \text{NaNO}_3 + \ \text{Al}_2\text{S}_3\end{align*}
 how many grams of aluminum sulfide are produced from \begin{align*}3.25 \ \mathrm{g}\end{align*} of aluminum nitrate?
 how many grams of sodium sulfide are required to produce \begin{align*}18.25 \ \mathrm{g}\end{align*} of aluminum sulfide?
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