What is stoichiometry?

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

$$2 \, H_2 + O_2 \longrightarrow 2 \, H_2O$$

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!

Example 1 – Space shuttle fuel mixture

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

$$2 \, H_2 + O_2 \longrightarrow 2 \, H_2O$$

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 O₂ that we have:

$$ \begin{align} \require{cancel} 629,340 \; \cancel{Kg \; O_2} &\left( \frac{1000 \; \cancel{g \; O_2}}{1 \; \cancel{Kg \; O_2}} \right)\left( \frac{1 \; mol \; O_2}{32 \; \cancel{g \; O_2}} \right) \\[5pt] &= 1.97 \times 10^7 \; mol \; O_2 \end{align}$$

$$1.97 \times 10^7 \; \cancel{mol \; O_2} \left( \frac{2 \cancel{mol \; H_2}}{1 \; \cancel{mol \; O_2}} \right) \left( \frac{2 \; \cancel{g \; H_2}}{1 \; \cancel{mol \; H_2}} \right) \left( \frac{1 \; Kg \; H_2}{1000 \; \cancel{g \; H_2}} \right) = 7.9 \times 10^4 \; Kg \; H_2$$

The first parentheses is the mole ratio: two H₂ for every one O₂. The second is the formula weight of H₂, and the third converts from grams to Kilograms.

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 H₂.

So about 79,000 Kg of H₂ would be needed to react with 629,340 Kg of LOX.

Example 2

How many moles of nitric acid (HNO₃) are required to completely dissolve 1 g of copper (Cu) according to the reaction

$$2 \, HNO_{3 \; (aq)}+Cu_{(s)} \rightleftharpoons CuNO_{3 \; (aq)}+NO_{2 \; (g)}+H_2O_{(l)}$$

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:

$$1 \; \cancel{g \; Cu} \left( \frac{1 \; mol \; Cu}{63.54 \; \cancel{g \; Cu}} \right) = 1.57 \times 10^{-2} \; mol \; Cu$$

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 HNO₃. That's it!

We could, of course, convert the number of moles of HNO₃ to grams and then even to milliliters of HNO₃ solution if we know its concentration.

Example 3

In the above example, how many grams of a 30% (by weight) aqueous solution of HNO₃ 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 HNO₃, then deal with that percent part:

$$ \begin{align} 3.64 \times 10^{-2} \; \cancel{mol \; HNO_3} &\left( \frac{63 \; g \; HNO_3}{1 \; \cancel{mol \; HNO_3}} \right) \\[5pt] &= 2.29 \; g \; HNO_3 \end{align}$$

That's how many grams of HNO₃ we need, now we need to find out how much of the 30% solution that translates to.The calculation we want is:

Example 4

n-octane (C₈H₁₈) is a principal component of gasoline. It oxidizes (burns) in the presence of oxygen according to the equation

$$2 \, C_8H_{18 \; (g)}+25 \, O_{2 \;(g)} \rightleftharpoons 16 \, CO_{2 \; (g)} + 18 \, H_2O_{(g)}$$

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 CO₂ 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 CO₂ 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 C₈H₁₈ to CO₂. 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 C₈H₁₈. 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:

$$ \begin{align} 3.785 \; \cancel{\text{liters }C_8H_{18}} &\left( \frac{750 \; g \; C_8H_{18}}{1 \; \cancel{\text{liter }C_8H_{18}}} \right) \\[5pt] &= 2.838 \times 10^3 \; g \; C_8H_{18} \end{align}$$

$$1 \; \cancel{gal \; C_8H_{18}} \left( \frac{3.785 \; \cancel{L}}{1 \cancel{gal}} \right)\left( \frac{750 \; \cancel{g \; C_8H_{18}}}{1 \; \cancel{L C_8H_{18}}} \right) \left( \frac{1 \; \cancel{mol \; C_8H_18}}{114 \; \cancel{g \; C_8H_{18}}} \right) \left( \frac{16 \; \cancel{mol \; CO_2}}{2 \; \cancel{mol \; C_8H_{18}}} \right) \left( \frac{22.4 \; L \; CO_2}{1 \; \cancel{mol \; CO_2}} \right)$$

All of the units cancel nicely in a nice progression from gallons of octane to liters of CO₂. 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:

= 3.785 · 750 · 16 · 22.4 / (114 ·2) liters

Example 5 — Limiting reactant

Gold (Au) metal reacts with chlorine gas (Cl₂) to form gold (III) chloride, AuCl₃.

$$2 \, Au_{(s)} + 3 \, Cl_{2 \; (g)} \rightleftharpoons 2 \, AuCl_{3 \; (g)}$$

How many grams of AuCl₃ will be formed if 1 g of gold metal is reacted with 2 g of Cl₂ ?

Solution: Problems like these are tricky. It's unlikely that 1 g of Au and 2 g of Cl₂ 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:

$$ \begin{align} 1.0 \; \cancel{g \; Au} \left( \frac{1 \; mol \; Au}{197 \; \cancel{g \; Au}} \right) &= 0.005 \; mol \; Au \\[5pt] g.0 \; \cancel{g \; Cl_2} \left( \frac{1 \; mol \; Cl_2}{70.97 \; \cancel{g \; Cl_2}} \right) &= 0.028 \; mol \; Cl_2 \end{align}$$

Now look again at the reaction:

$$2 \, Au_{(s)} + 3 \, Cl_{2 \; (g)} \rightleftharpoons 2 \, AuCl_{3 \; (g)}$$

Example 6 — Limiting reactant

152 grams of carbon monoxide (CO) are added to 24 g of hydrogen gas (H₂) to produce methanol (CH₃OH). How many grams of methanol* will be produced?

*Methanol is also called methyl alcohol

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

$$CO + H_2 \longrightarrow CH_3OH$$

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

$$CO + 2 \, H_2 \longrightarrow CH_3OH$$

Now we just need to calculate the number of moles of CH₃OH we could expect if 24 g of H₂ reacted completely (assuming excess CO) and if 152 g of CO reacted completely (assuming excess H₂). The lesser number of moles will be the theoretical maximum number of moles of CH₃OH formed. Here are the calculations:

$$ \begin{align} 24 \; \cancel{g \; J_2} &\left( \frac{1 \; mol \; H_2}{2 \; \cancel{g \; H_2}} \right)\left( \frac{1 \; mol\; CH_3OH}{2 \; \cancel{mol \; H_2}} \right) \\[5pt] &=6 \; mol \; CH_3OH \\[5pt] 152 \; \cancel{g \; CO} &\left( \frac{1 \; mol \; CO}{28 \; \cancel{g \; CO}} \right)\left( \frac{1 \; mol \; CH_3OH}{1 \; \cancel{mol \; CO}} \right) \\[5pt] &= 5.42 \; mol \; CH_3OH \end{align}$$

Example 7 — One more limiting reactant problem ...

93 Kg of nitrogen gas (N₂) are added to 265.8 Kg of hydrogen gas (H₂) to produce ammonia gas (NH₃). How much ammonia (in Kg) can be produced by this reaction?

Solution: First, the balanced reaction is

$$N_2 + 3\, H_2 \rightleftharpoons 2 \, NH_3$$

Now we calculate the number of moles of NH₃ we could expect if 93 Kg of N₂ reacted completely (assuming excess H₂) and if 265.8 Kg of H₂ reacted completely (assuming excess N₂). The lesser number of moles will be the theoretical maximum number of moles of NH₃ formed. First the N₂:

$$ \begin{align} 93,333 \; \cancel{g \; N_2} &\left( \frac{1 \; \cancel{mol \; N_2}}{28 \; \cancel{g \; N_2}} \right) \left( \frac{2 \; mol \; NH_3}{1 \; \cancel{mol \; N_2}} \right) \\[5pt] &= 6,643 \; mol \; NH_3 \end{align}$$

Then the H₂:

Practice problems

$\longleftarrow$ To view wide equations in solutions, scroll left | right. $\longrightarrow$

1. Consider the reaction

   $N_{2 \; (g)} + F_{2 \; (g)} \longrightarrow NF_{3 \; (g)}$

   Balance the equation if necessary. If 27.5 g of nitrogen gas (N₂) are used in the reaction, what mass of fluorine (F₂) would be needed for the reaction to go to completion?

   Solution

   The balanced equation is: $N_2 + 3 \, F_2 \longrightarrow 2 \, NF_3$.

   $$27.5 \; \cancel{g \; N_2} \left( \frac{1 \; \cancel{mol \; N_2}}{28 \; \cancel{g \; N_2}} \right) \left( \frac{3 \; \cancel{mol \; F_2}}{1 \; \cancel{mol \; N_2}} \right) \left( \frac{38 \; g \; F_2}{1 \; \cancel{mol \; F_2}} \right) = \frac{27.5 \cdot 3 \cdot 38}{28} = 112 \; g \; F_2$$

   The formula weight of N₂ is 28 g/mol (2 × 14, the atomic weight of nitrogen) and the FW of F₂ is 38 g/mol (2 × 19, the atomic weight of fluorine).




2. Consider the reaction

   $C_6H_{10 \; (g)} + O_{2 \; (g)} \longrightarrow CO_{2 \; (g)} + H_2O_{(g)}$

   Balance the equation if necessary. If only 18.3 g of O₂ is available, how much C₆H₁₀ (cyclohexene) can be reacted? How much CO₂ (in grams) would be produced by this reaction?

   Solution

   The balanced equation is: $2 \, C_6H_{10} + 17 \, O_2 \longrightarrow 12 \, CO_2 + 10 \, H_2O$

   $$ \begin{align} 18.3 \; \cancel{g \; O_2} \left( \frac{1 \; \cancel{mol \; O_2}}{32 \; \cancel{g \; O_2}} \right) \left( \frac{12 \; \cancel{mol \; CO_2}}{17 \; \cancel{mol \; O_2}} \right) \left( \frac{44 \; g \; CO_2}{1 \; \cancel{mol \; CO_2}} \right) &= \frac{18.3 \cdot 12 \cdot 44}{32 \cdot 17} \\[5pt] &= 17.8 \; g \; CO_2 \end{align}$$

   This is a combustion reaction.




3. Consider the reaction

   $HBr_{(aq)} + KHCO_{3 \; (aq)} \longrightarrow H_2O_{(l)} + KBr_{(aq)} + CO_{2 \; (g)}$

   1. Balance the reaction. What type of reaction is this?
   2. How many moles of KBr would be formed by reacting 3.3 moles of HBr with plenty of KHCO₃?
   3. How many grams of KBr would be formed by reacting 3.3 moles of HBr with only 1.2 moles of KHCO₃?
   Solution

   (a)   The equation is balanced.   (b) ↓

   $$3.33 \; \cancel{mol \; HBr} \left( \frac{1 \; mol \; KBr}{1 \; \cancel{mol \; HBr}} \right) = 3.3 \; mol \; KBr$$

   (c)   In this reaction HBr and KHCO₃ are in a 1:1 ratio, so 1.2 moles of KHCO₃ is the limiting reactant:

   $$1.2 \; \cancel{mol \; KHCO_3} \left( \frac{1 \; \cancel{mol \; KBr}}{1 \; \cancel{mol \; KHCO_3}} \right) \left( \frac{119 \; g \; KBr}{1 \; \cancel{mol \; Kbr}} \right) = 143 \; g \; KBr$$

   N.B. This reaction is difficult to classify. It's part displacement and part decomposition (more products than reactants). Many reactions don't fit neatly into any one category.




4. In 2018, citizens and businesses of the US burned about 29.9 trillion cubic feet of natural gas. That's roughly 1.6 × 10⁹ Kg of propane, if we assume that natural gas is mostly propane (not a terrible assumption, but it also contains methane, butane and other small hydrocarbon gases and contaminants).

   1. Write a balanced reaction for the combustion of propane, C₃H₈.
   2. Assuming plenty of oxygen is available for the reactions, calculate the number of metric tons (1 metric ton = 1000 Kg) of CO₂ that are released into the atmosphere by these reactions.
   Solution

   The balanced equation is

   $$C_3H_8 + 5 \, O_2 \longrightarrow 3 \, CO_2 + 4 \, H_2O$$

   $$ \begin{align} 1.6 \times 10^9 \; \cancel{Kg \; C_3H_8} &\left( \frac{1 \; \cancel{mol \; C_3H_8}}{0.044 \; \cancel{Kg \; C_3H_8}} \right) \left( \frac{3 \; \cancel{mol \; CO_2}}{1 \; \cancel{mol \; C_3H_8}} \right) \left( \frac{0.044 \; Kg \; CO_2}{1 \; \cancel{mol \; CO_2}} \right) \\[5pt] &= 4.8 \times 10^6 \text{ metric tons of } CO_2 \end{align}$$

   Notice that the formula weights (FW) of C₃H₈ and CO₂ are coincidentally the same.




5. Consider the reaction:

   $KClO_{3 \; (aq)} \longrightarrow KCl_{(aq)} + O_{2 \; (g)}$

   1. Balance the reaction. What type of reaction is this?
   2. If 19.5 g of oxygen gas (O₂) are formed by this reaction, calculate the amount of KClO₃ that must have been present. Assume that the reaction goes to completion. (By the way, it is pretty rare for a reaction to actually go to completion.)
   Solution

   The balanced equation (a decomposition) is:

   $$2 \, KClO_3 \longrightarrow 2 \, KCl + 3 \, O_2$$

   The stoichiometry is

   $$28 \; \cancel{g \; O_2} \left( \frac{1 \; \cancel{mol \; O_2}}{34 \; \cancel{g \; O_2}} \right) \left( \frac{2 \; \cancel{mol \; KClO_3}}{3 \; \cancel{mol \; O_2}} \right) \left( \frac{122.55 \; g \; KClO_3}{2 \; \cancel{mol \; KClO_3}} \right) = 49.8 \; g \; Cl_2$$




6. Consider the reaction

   $K_{(s)} + Cl_{2 \; (ag)} \longrightarrow KCl_{(s)}$

   1. Balance the reaction. What type of reaction is this?
   2. If 1.5 g of potassium (K) reacts with 194 g of chlorine gas (Cl₂), how much KCl will be formed?
   3. If 28 g of KCl were formed in this reaction, how much Cl₂ (in grams) must have reacted?
   Solution

   This is a synthesis reaction:

   $$2 \, K + Cl_2 \longrightarrow 2 \, KCl$$

   Limiting reactant: In this reaction we need twice as many moles of K as Cl₂ and we don't have nearly that many (see calculations below).

   $$ \begin{align} 1.5 \; \cancel{g \; K} \left( \frac{1 \; mol \; K}{39.1 \; \cancel{g \; K}} \right) = 0.038 \; mol \; K \\[5pt] 194 \; \cancel{g \; Cl_2} \left( \frac{1 \; mol \; Cl_2}{70.9 \; \cancel{g \; Cl_2}} \right) = 2.73 \; mol \; Cl_2 \end{align}$$

   So K is the limiting reactant. Now we'll base all further work on the 0.038 moles of K.

   $$\text{(b)} \; \; 0.038 \; mol \; K \left( \frac{2 \; \cancel{mol \; KCl}}{2 \; \cancel{mol \; K}} \right) \left( \frac{74.55 \; \cancel{g \; KCl}}{1 \; \cancel{mol \; KCl}} \right) = 2.83 \; g \; KCl$$

   $$\text{(c)} \; \; 28 \; \cancel{g \; KCl} \left( \frac{1 \; \cancel{mol \; KCl}}{74.55 \; \cancel{g \; KCl}} \right) \left( \frac{1 \; \cancel{mol \; Cl_2}}{2 \; \cancel{mol \; KCl}} \right) \left( \frac{70.9 \; g \; Cl_2}{1 \; \cancel{mol \; Cl_2}} \right) = 13.3 \; g \; Cl_2$$




7. Consider the reaction:

   $Cu_{(s)} + AgNO_{3 \; (aq)} \longrightarrow Cu(NO_3)_{2 \; (aq)} + Ag_{(s)}$

   1. Balance the reaction. What type of reaction is this?
   2. How many grams of copper (Cu) would be required, assuming plenty of AgNO₃ to react with, to form 2.8 g of silver metal (Ag)?
   3. If 1.0 g of AgNO₃ was reacted with 92.0 g of Cu metal (copper), how much Cu(NO₃)₂ would be formed in this reaction?
   Solution

   $Cu + 2 \, AgNO_3 \longrightarrow Cu(NO_3)_2 + 2 \, Ag$   (single displacement)

   $$\text{(b)} \; 2.8 \; \cancel{g \; Ag} \left( \frac{1 \; \cancel{mol \; Ag}}{107.9 \; \cancel{g \; Ag}} \right) \left( \frac{1 \; \cancel{mol \; Cu}}{2 \; \cancel{mol \; Ag}} \right) \left( \frac{63.54 \; g \; Cu}{1 \; \cancel{mol \; Cu}} \right) = 0.82 \; g \; Cu$$

   Now find the limiting reactant:

   $$ \begin{align} 1.0 \; \cancel{g \; AgNO_3} \left( \frac{1 \; mol \; AgNO_3}{169.9 \; \cancel{g \; AgNO_3}} \right) &= 0.0059 \; mol \; AgNO_3 \\[5pt] 92 \; \cancel{g \; Cl_2} \left( \frac{1 \; mol \; Cu}{63.54 \; \cancel{g \; Cu}} \right) &= 1.45 \; mol \; Cu \end{align}$$

   In this reaction we need twice as many moles of AgNO₃ as Cu and we don't have nearly that many. Therefore AgNO₃ is the limiting reactant. Now base all further work on 0.0059 moles of AgNO₃.

   $$\text{(c)} \; 0.0059 \; \cancel{mol \; AgNO_3} \left( \frac{1 \; \cancel{mol \; Cu(NO_3)_2}}{2 \; \cancel{mol \; AgNO_3}} \right) \left( \frac{187.65 \; g \; Cu(NO_3)_2}{1 \; \cancel{mol \; Cu(NO_3)_2}} \right) = 0.55 \; g \; Cu(NO_3)_2$$




8. Methyl mercaptan (CH₄S or CH₃SH) is one of the stinky (rotten egg smell) gases added to household natural gas so that humans can detect leaks before being asphyxiated by them. The combustion of methyl mercaptan produces sulfur dioxide:

   $CH_4S + 3 \, O_2 \longrightarrow CO_2 + 2 \, H_2O + SO_2$

   1. How many grams of sulfur dioxide (SO₂) are formed from complete combustion of 5 g of CH₄S ?
   2. How many grams of SO₂ are formed when 5 g of CH₄S react with 2.0 g of O₂ ?
   Solution

   $CH_4S + 3 \; O_2 \longrightarrow CO_2 + 2 \, H_2O + SO_2$   (combusion of a C,H,S compound)

   $$\text{(a)} \; 5 \; \cancel{g \; CH_4S} \left( \frac{1 \; \cancel{mol \; CH_4S}}{48 \; \cancel{g \; CH_4S}} \right) \left( \frac{1 \; \cancel{mol \; SO_2}}{1 \; \cancel{mol \; CH_4S}} \right) \left( \frac{64 \; g \; SO_2}{1 \; \cancel{mol \; SO_2}} \right) = 6.67 \; g \; SO_2$$

   (b) Here we need to find the limiting reactants:

   $$ \begin{align} 5.0 \; \cancel{g \; CH_4S} \left( \frac{1 \; mol \; CH_4S}{48 \; \cancel{g \; CH_4S}} \right) = 0.1042 \; mol \; CH_4S \\[5pt] 2.0 \; \cancel{g \; O_2} \left( \frac{1 \; mol \; Cu}{32 \; \cancel{g \; O_2}} \right) = .0625 \; mol \; O_2 \end{align}$$

   In this reaction we need three times as many moles of O₂ as CH₄S, and we don't have nearly that much, therefore O₂ is the limiting reactant. Now base all further work on 0.0625 moles of O₂.

   $$0.0625 \; \cancel{mol \; O_2} \left( \frac{1 \; \cancel{mol \; SO_2}}{3 \; \cancel{mol \; O_2}} \right) \left( \frac{64 \; g \; SO_2}{1 \; \cancel{mol \; SO_2}} \right) = 1.33 \; g \; SO_2$$




9. Consider the reaction:

   $Au_2S_{3 \; (aq)} + H_{2 \; (g)} \longrightarrow H_2S_{(g)} + Au_{(s)}$

   If 500.2 g of Au₂S₃ (gold sulfide) and 5.67 g of H₂ react, calculate the amount (in grams) of each of the products.

   Solution

   Balanced reaction: $Au_2S_3 + 3 \, H_2 \longrightarrow 3 \, H_2S + 2 \, Au$

   First determine the limiting reactant:

   $$ \begin{align} 500.2 \; \cancel{g \; Au_2S_3} \left( \frac{1 \; mol \; Au_2S_3}{426 \; \cancel{g \; Au_2S_3}} \right) = 1.174 \; mol \; Au_2S_3 \\[5pt] 5.67 \; \cancel{mol \; H_2} \left( \frac{1 \; mol \; H_2}{2 \; \cancel{g \; H_2}} \right) = 2.835 \; mol \; H_2 \end{align}$$

   In this reaction we need three times as many moles of H₂ as Au₂S₃, and we don't have quite that many, so H₂ is the limiting reactant. Now base all further calculations on 2.835 moles of H₂

   $$ \begin{align} 2.835 \; \cancel{mol \; H_2} \left( \frac{3 \; \cancel{mol \; H_2S}}{3 \; \cancel{mol \; H_2}} \right) \left( \frac{34.1 \; g \; H_2S}{1 \; \cancel{mol \; H_2S}} \right) = 96.7 \; g \; H_2S \\[5pt] 2.835 \; \cancel{mol \; H_2} \left( \frac{2 \; \cancel{mol \; Au}}{3 \; \cancel{mol \; H_2}} \right) \left( \frac{197 \; g \; Au}{1 \; \cancel{mol \; Au}} \right) = 372 \; g \; Au \end{align}$$




10. Consider the reaction:

    $Mg_3N_2 + 6 \, H_2O_{(l)} \longrightarrow 3 \, Mg(OH)_{2 \; (aq)} + 2 \, NH_{3 \; (aq)}$

    If 58.1 g of Mg₃N₂ and 20.4 g of H₂O react, calculate the masses of each of the products.

    Solution

    Balanced: $Mg_3N_2 + 6 \, H_2O \longrightarrow 3 \, Mg(OH)_2 + 2 \, NH_3$

    Find the limiting reactant:

    $$ \begin{align} 51.8 \; \cancel{g \; Mg_3N_2} \left( \frac{1 \; mol \; Mg_3N_2}{101 \; \cancel{g \; Mg_3N_2}} \right) = 0.513 \; mol \; Mg_3N_2 \\[5pt] 20.4 \; \cancel{g \; H_2O} \left( \frac{1 \; mol \; H_2O}{2 \; \cancel{g \; H_2O}} \right) = 1.133 \; mol \; H_2O \end{align}$$

    In this reaction we need six times as many moles of water as Mg₃N₂, and we have about half that much, so H₂O is the limiting reactant. Now base all further calculations on 1.133 moles of H₂O.

    $$ \begin{align} 1.133 \; \cancel{mol \; H_2O} \left( \frac{3 \; \cancel{mol \; Mg(OH)_2}}{6 \; \cancel{mol \; H_2O}} \right) \left( \frac{58.3 \; g \; Mg(OH)_2}{1 \; \cancel{mol \; Mg(OH)_2}} \right) &= 33.0 \; g \; Mg(OH)_2 \\[5pt] 1.133 \; \cancel{mol \; H_2O} \left( \frac{2 \; \cancel{mol \; NH_3}}{6 \; \cancel{mol \; H_2O}} \right) \left( \frac{17 \; g \; NH_3}{1 \; \cancel{mol \; NH_3}} \right) &= 6.42 \; g \; NH_3 \end{align}$$

Video examples

1. Cu_(s) + AgNO_{3 (aq)} → Cu(NO₃)_{2 (aq)} + Ag_(s)

How many grams of copper metal (assuming excess AgNO₃) would it take to form 2.8 g of silver metal, assuming that the reaction proceeded to completion? When you see language like "excess AgNO₃" you can assume that this is not a limiting reactant problem.

Minutes of your life: 2:59

3. A limiting reactant problem

8 g of gold (Au) react with 1.1. liters of Cl₂ gas at 2 atm. pressure (T = 250˚C) to produce gold (III) chloride. How much AuCl₃ can possibly be produced by this reaction?

In this problem, you'll have to write a balanced reaction first. The number of moles of Cl₂ 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