When the space shuttle took off it used two different rocket systems. The main engines were powered by combining **oxygen** and **hydrogen** in the highly exothermic reaction

The fuel was stored in that brown tank in liquid form as liquid oxygen (LOX) and liquid hydrogen (LH2). Liquids are much more dense than gases, so storage in the liquid phase allows the shuttle to carry an immense amount of fuel.

Unfortunately, that fuel made up the majority of the *weight* of the shuttle at liftoff. So most of what the shuttle engines lifted from the ground was fuel. That means it was very important not to have too much of one or the other reactant on board. We'd want *just enough* hydrogen to react with the amount of oxygen on board, and vice versa. That's where **stoichiometry** and moles are essential.

And maybe you can be the one to invent a kind of rocket propulsion that will release more energy per pound of fuel!

*Space Shuttle launch – NASA*

**Stoichiometry** is the relationship between the relative quantities of substances involved in a chemical reaction, typically ratios of whole numbers. Stoichiometry is a catch all word for all of the calculations we do to determine how much of what to mix with what.

Let's begin with a calculation relevant to our opening example: The space shuttle LOX tank held 629,340 Kg (about 630 metric tons) of LOX. **How many kilograms of Hydrogen would it need to carry in order to assure that the reaction**

goes to completion, with no reactants remaining?** (The backward rate of this reaction is **extremely slow compared to the explosive forward rate, thus the single arrow)

__Solution__: We start by determining the number of moles of O_{2} that we have:

In the calculation above, I've combined the conversion from Kg to grams with the calculation of the number of moles of O_{2} (32 g/mol). That's easy if we keep track of units. Now that we have the number of moles of O_{2}, we use the balanced chemical equation (that's what it's for!) to find out how many moles of H_{2} we need. From the coefficients, you can see that *two* moles of H_{2} are consumed for every one mole of O_{2}, so we need 3.94 x 10^{7} mol of H_{2} (1.97 x 2 = 3.94)

The **mole ratio** told us that in this reaction exactly twice the number of moles of H_{2} are used as O_{2}. Now it's just a simple matter of converting 3.94 x 10^{4} mol of H_{2} to Kg of H_{2}.

We could move forward and complete the calculation one step at a time - moles O_{2} to moles H_{2} to grams H_{2} to Kg H_{2}, but there's a faster way if we make good use of units and cancellation of units. Here's how to combine the second part of the calculation into one big step:

Notice that in each of the large fractions above, a relationship and its units are written, and written in such a way that the units cancel with those of the previous term to get us closer to the desired units, in this case Kg of H_{2}.

So about 79,000 Kg of H_{2} would be needed to react with 629,340 Kg of LOX.

It turns out that in practice the shuttles carried a little *more* LH2 than that because of engine efficiency issues and because the rate of loss of lighter LH2 due to evaporation during the wait for launch is higher than that of LOX.

Nevertheless, I hope you get the idea. This is a very important calculation and it depends 100% on doing good stoichiometry.

__Solution__: The first thing to do is to convert the problem to the common currency of all stoichiometry problems, the mole. We need to figure out how many moles of Cu are in 1 g of Cu:

Now that we've got the number of moles of copper, the balanced equation (check it!) tells us that for every one mole of Cu, 2 moles of nitric acid are needed, so we need 3.64 x 10^{-2} moles of HNO_{3}. That's it!

We could, of course, convert the number of moles of HNO_{3} to grams and then even to milliliters of HNO_{3} solution if we know its concentration.

Source: Education in Chemistry, Nov. 2011

In the above example, how many grams of a 30% (by weight) aqueous solution of HNO_{3} would be required to run the reaction to completion?

__Solution__: In this situation, we already know that we need 3.64 x 10^{-2} moles of nitric acid, so we just need to convert to grams of HNO_{3}, then deal with that percent part:

That's how many grams of HNO_{3} we need, now we need to find out how much of the 30% solution that translates to.The calculation we want is:

Often (particularly in the biological sciences) solution concentrations are given in **percent**, either weight of solute divided by total weight of solution %(w/w) or weight of solute over total volume of solution %(w/v).

See the notes on concentration to brush up on how to specify the concentration of solutions.

If 1 gallon of gasoline is burned in the presence of excess oxygen (i.e. enough oxygen to make the reaction go to completion), how many liters of CO_{2} gas are emitted into the atmosphere? There are 3.785 liters in a gallon and the density of gasoline is about 0.75 Kg/liter. At atmospheric temperature and pressure, 1 mole of CO_{2} gas takes up about 22.4 liters of volume.

Photo: California Air Resources Board

__Solution__: First, look at the balanced equation (and make sure it's balanced). This question only asks us to compare amounts of C_{8}H_{18} to CO_{2}. It says that there will be more than enough oxygen, and we don't really care how much water is produced. So we begin by finding the number of moles of C_{8}H_{18}. First we need to convert from gallons to liters using the conversion given: **1 gallon = 3.785 liters**.

Now we can use the density of gasoline to find the approximate mass of octane (we're assuming that gas is 100% octane here, not too bad an approximation).

Here's the mass of octane:

Now convert to moles:

Now we use the mole ratio in the balanced equation: For every two moles of C_{8}H_{18} we form 16 moles of CO_{2}.

Now we can use the molar volume of CO_{2} gas (the volume that one mole takes up - given) to find our result:

There are 1000 liters in a cubic meter, so that's about 4.5 cubic meters of CO_{2} gas emitted from every gallon of gas burnt. Something to think about.

Now we could easily have done this whole calculation at once by just multiplying the many terms in parenthesis above and canceling units. Here's how it looks:

All of the units cancel nicely in a nice progression from gallons of octane to liters of CO_{2}. Remember that division is just multiplication by the reciprocal and that multiplication is commutative, so these numbers can be multiplied and divided on a calculator in any order, like:

**PS**: When you study gases, you'll learn that a mole of any gas at standard temperature and pressure, T = 273K and P = 1 atm, occupies 22.4 liters. That was **Avogadro's** original conjecture.

How many grams of AuCl_{3} will be formed if 1 g of gold metal is reacted with 2 g of Cl_{2} ?

__Solution__: Problems like these are tricky. It's unlikely that 1 g of Au and 2 g of Cl_{2} are exactly "**stoichiometric**." That is, it's unlikely that there's just enough of both so that at the end of the reaction, all reactants are used up and we only have products.

That means that one of the reactants will run out first, and we call this the **limiting reactant**. When it runs out, the reaction has to stop. So our first job in a problem like this is to find out which reactant is limiting.

We start by calculating the number of moles of each reactant present:

Now look again at the reaction:

**2 Au _{(s)} + 3 Cl_{2 (g)} ⇌ 2 AuCl_{3 (g)}**

It says that for every 2 moles of gold, we need 3 moles of chlorine gas. In fact, it's easy to see that we have more Cl_{2} than that. Look at it the other way: The equation says that for every three moles of Cl_{2}, we need 2 moles of gold. It's not even close. Either way we look at it, **Au is the limiting reactant**.

Now that we've identified the limiting reactant, the trick is to base all further calculations on it, because its amount alone determines how far the reaction will proceed. Now it's just a matter of finding out how many moles of AuCl_{3} will be formed:

(We could have done that just by inspection, of course!). Finally, the mass of AuCl_{3} produced:

**Methanol is also called methyl alcohol*

__Solution__: First we ought to write a reaction. It's just an addition reaction,

**CO + H _{2} → CH_{3}OH**

Well, that's an unbalanced reaction, so we ought to balance it. That turns out to be simple:

**CO + 2 H _{2} → CH_{3}OH**

Now we just need to calculate the number of moles of CH_{3}OH we could expect if 24 g of H_{2} reacted completely (assuming excess CO) and if 152 g of CO reacted completely (assuming excess H_{2}). The lesser number of moles will be the theoretical maximum number of moles of CH_{3}OH formed. First the H_{2}:

Then the CO:

Now it's easy to see that the limiting reagent is CO. When 5.4 moles of methanol has been produced, the supply of CO will be exhausted, with 0.6 moles of H_{2} remaining. Now we just need to calculate the mass of CH_{3}OH that is 5.4 moles:

To solve a **limiting reactant** problem, calculate the number of moles of the desired product that would be obtained from each of the reactants, assuming an **excess** of all other reactants. The reactant that yields the **least** product is the limiting reactant, and that yield of product is the **theoretical maximum yield** of the reaction.

__Solution__: First, the balanced reaction is

**N _{2} + 3 H_{2} → 2 NH_{3}**

Now we calculate the number of moles of NH_{3} we could expect if 93 Kg of N_{2} reacted completely (assuming excess H_{2}) and if 265.8 Kg of H_{2} reacted completely (assuming excess N_{2}). The lesser number of moles will be the theoretical maximum number of moles of NH_{3} formed. First the N_{2}:

Then the H_{2}:

Well, it's very easy to see that the limiting reagent is N_{2}. When 6,643 moles of ammonia has been produced, the supply of N_{2} will be exhausted, with many more moles of H_{2} remaining. Now we just need to calculate the mass of NH_{3} that constitutes 6,643 moles:

You will recall from your study of bonding that the triple bond in N**-**N is very strong. In fact, there are only a couple of things in nature that can break it. One is an energetic electric arc like in lightning. Another very important one is the biochemistry of plants called **legumes**. These plants can break the N-N bond to form atomic nitrogen thats available for use in making amino acids and nucleic acids. We couldn't get along without them.

In modern agriculture, it's very important to have non-N_{2} nitrogen available in the soil, but that supply gets depleted over time as we grow more food on the land, so we need to add nitrogen-rich fertilizer to it. The primary source of that nitrogen is **fertilizer**. The reaction to synthesize ammonia is run using a catalyst and the process is called the **Haber process**. It's one of the most important chemical reactions we have.

How many grams of copper metal (assuming excess AgNO_{3}) would it take to form 2.8 g of silver metal, assuming that the reaction proceeded to completion? When you see language like "excess AgNO_{3}" you can assume that this is not a limiting reactant problem.

Minutes of your life: 2:59

If 19.5 g of oxygen (O_{2}) are formed by this reaction, calculate the amount of KClO_{3} that must have been initially present. Assume that the reaction goes to completion.

Minutes of your life: 2:20

8 g of gold (Au) react with 1.1. liters of Cl_{2} gas at 2 atm. pressure (T = 250˚C) to produce gold (III) chloride. How much AuCl_{3} can possibly be produced by this reaction?

In this problem, you'll have to write a balanced reaction first. The number of moles of Cl_{2} is determined using the ideal gas law: n = PV/RT. Notice that this is a **limiting reactant problem **because we have fixed amounts of all reactants.

Minutes of your life: 5:18

If 500.2 g of Au_{2}S_{3} and 5.67 g of H_{2} react, calculate the amount (in grams) of gold metal that will be formed. Notice that this is a **limiting-reactant problem**: We know the amounts of both reactants,and it's up to us to determine the maximum amount of product that can be obtained from that mixture.

Minutes of your life: 5:38

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